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Full file at https://fratstock.euProblem 2.1 The magnitudes |FA| = 60 N and|FB | = 80 N. The angle α = 45◦. Graphically de-termine the magnitude of the sum of the forces F =FA + FB and the angle between FB and F.Strategy: Construct the parallelogram for determiningthe sum of the forces, drawing the lengths of FA and FB
proportional to their magnitudes and accurately measur-ing the angle α, as we did in Example 2.1. Then youcan measure the magnitude of their sum and the anglebetween their sum and FB .
FA
FB
Solution: The graphical construction is shown:The angle β is graphically determined to be about 26◦ and the angleθ is about θ = 19◦. The magnitude of the sum |F| = |FA + FB | isabout |F| = 130 N. (These values check with a determination using
trigonometry, β = 25.9◦, θ = 19.1◦, and |F| = 129.6 N .) 45˚
FA
FB
F
Problem 2.2 The magnitudes |FA| = 40 N and|FA + FB | = 80 N. The angle α = 60◦. Graphicallydetermine the magnitude of FB .
Solution: Measuring, FB∼= 52 N
FA
FB
60°
0 40 50 80 100 N
FA
FA + FB
FB
Full file at https://fratstock.euProblem 2.3 The magnitudes |FA| = 100 lb and|FB | = 140 lb. The angle α = 40◦. Use trigonom-etry to determine the magnitude of the sum of the forcesF = FA + FB and the angle between FB and F.Strategy: Use the laws of sines and cosines to analyzethe triangles formed by the parallelogram rule for thesum of the forces as we did in Example 2.1. The laws ofsines and cosines are given in Section A.2 of Appendix A.
Solution: The construction is shown. Use the cosine law to de-termine the magnitude |F|.
|F|2 = |FA|2 + |FB |2 − 2|FA||FB | cos 140◦
|F|2 = (100)2 + (140)2 − 2 (100) (140) cos 140◦
or |F| =√
5.1049 . . . (104) = 225.94 . . . = 225.9 lb
Use the law of sines to determine the angle θ., i.e., |F|sin 140◦ = |FB |
sin θ.
From which we get sin θ = |FB ||F| sin 140◦ = 0.398, and θ =
23.47◦. The angle between FB and F is
Thus β = 40 − 23.47 = 16.53◦
40°140FA
FB
F
Problem 2.4 The magnitudes |FA| = 40 N and |FA+FB | = 80 N. The angle α = 60◦. Use trigonometry todetermine the magnitude of FB .
Solution: Draw the force triangle.From the law of sines
|FA + FB |sin 120◦ =
|FA|sin θ
sin θ =|FA|
|FA + FB | sin 120◦
sin θ =(
4080
)(0.866)
θ = 25.66◦
θ + γ + 120◦ = 180◦ ⇒ γ = 34.34◦
From the law of sines,
|FB |sin γ
=|FA + FB |sin 120◦
|FB | =(
sin γ
sin 120◦
)(80 N)
|FB | = 52.1 N
FA
FB
60°120°
FA
FB
F A F B
Full file at https://fratstock.euProblem 2.5 The magnitudes |FA| = 100 lb and|FB | = 140 lb. If α can have any value, what are theminimum and maximum possible values of the magni-tude of the sum of the forces F = FA + FB , and whatare the corresponding values of α?
Solution: A graphical construction shows that the magnitude isa minimum when the two force vectors are opposed, and a maximumwhen both act in the same direction. The corresponding values are
|F|max = |FA + FB | = |100 + 140| = 240 lb, and α = 0◦.
|F|min = |FA + FB | = |100 − 140| = 40 lb, and α = 180◦.
Problem 2.6 The angle θ = 30◦. What is the magni-tude of the vector rAC?
60 mm 150 mm
A
C
BrAB
rBC
rAC
Solution:
A30°
60 mm 150 mm
B
rACC
From the law of sines
BC
sin 30◦ =AB
sin α=
AC
sin γ
We know BC and AB. Thus
150sin 30◦ =
60sin α
⇒ α = 11.54◦
Also 30◦ + α + γ = 180◦ ⇒ γ = 138.46◦
Now, from the law of sines
150sin 30◦ =
AC
sin 138.46◦
AC = |rAC | = 199 mm
Full file at https://fratstock.euProblem 2.7 The vectors FA and FB represent theforces exerted on the pulley by the belt. Their magni-tudes are |FA| = 80 N and |FB | = 60 N. What is themagnitude |FA + FB | of the total force the belt exertson the pulley?
45°
FA
FB
10°
Solution:
3535°
45°145°45°
10° 10°
FB
FB
FA
F A + F B
FA
Law of cosines
|FA + FB | = (80)2 + (60)2 − 2(80)(60) cos 145◦
|FA + FB | = 133.66 ≈ 134 N
Law of sines
|FB |sin β
=|FA + FB |
sin 145⇒ 60
sin β=
133.66sin 145
β = 14.92◦
∴ |FA + FB | = 134 N
Problem 2.8 The magnitude of the vertical force F is80 N. If you resolve it into components FAB and FAC
that are parallel to the bars AB and AC, what are themagnitudes of the components?
B
C20°
30°
A
F
Solution: F is made up of components in the two known direc-tions. Since we also know the direction of F, we can draw a forcetriangle (FAB + FAC = F).Thus, we have a triangle DEF as shown
γ + 110◦ + 30◦ = 180◦
γ = 40◦
From the law of sines,
80sin 40◦ =
|FAC |sin 30◦ =
|FAB |sin 110◦
|FAB | = 117 N |FAC | = 62.2 N
60°
80 N
80 N110°
30°
30°
20°FAC
FD
E
FAC FAB
FAB
Full file at https://fratstock.euProblem 2.9 The rocket engine exerts an upward forceof 4 MN (meganewtons) magnitude on the test stand. Ifyou resolve the force into vector components parallel tothe bars AB and CD, what are the magnitudes of thecomponents? 30° 45°A
B D
C
Solution: The vector diagram construction is shown. From thelaw of sines,
|FAB |sin 45◦ =
|F|sin 75◦ ,
from which
|FAB | = |F|(
sin 45◦
sin 75◦
)= (4)(0.732) = 2.928 . . . = 2.93 MN
|FBC | = |F|(
sin 60sin 75
)= (4)(0.8966) = 3.586 . . . = 3.59 MN
30 45A
B D
C° °
A 38˚
68˚
45˚45˚
45˚45˚
C
D
F
BFBA FBC
Full file at https://fratstock.euProblem 2.10 If F is resolved into components paral-lel to the bars AB and BC, the magnitude of the compo-nent parallel to bar AB is 4 kN. What is the magnitudeof F?
CA
B
F
150 mm
100 mm400 mm
Solution:
CA
B
F
150 mm
100 mm400 mm
Call the force in AB FAB and the force in BC ∼ FBC . ThenFAB +FBC = F. We know the geometry and that |FAB | = 4 kN.Draw a diagram of the geometry, get the angles, then draw the forcetriangle.
C
B
A φθ500 mm
100 mm
150 mm
tan θ =150400
θ = 20.6◦
tan φ =150100
φ = 56.3◦
φ
α
γ
δ
θθ
FAB
FBC
F
FAB = 4 kN
γ + θ = 90◦ γ = 69.4◦
α + φ = 90◦ α = 33.7◦
δ + α + γ = 180◦ δ = 76.9◦
From the law of sines
|F|sin δ
=|FAB |sin α
|F| =[
sin(76.9◦)sin(33.7◦)
](4) = 7.02 kN
Full file at https://fratstock.euProblem 2.11 The forces acting on the sailplane arerepresented by three vectors. The lift L and drag D areperpendicular, the magnitude of the weightW is 3500 N,and W + L + D = 0. What are the magnitudes of thelift and drag?
W
D
L
25°
Solution: Draw the force triangle and then use the geometry plus
W
D
L
25°
cos 25◦ =|L||W|
sin 25◦ =|D||W|
|W| = 3500 N
|L| = 3500 cos 25◦
|D| = 3500 sin 25◦
|L| = 3170 N
|D| = 1480 N
25°65°
25°W
L
D
Problem 2.12 The suspended weight exerts a down-ward 2000-lb force F at A. If you resolve F into vectorcomponents parallel to the wires AB, AC, and AD, themagnitude of the component parallel to AC is 600 lb.What are the magnitudes of the components parallel toAB and AD?
A
B C D60° 70° 45°
Solution: We resolve the force exerted by the weight into com-ponents parallel to the wires:
FAD
FAC
FAB
45°
70°
60°
W
We see that
|FAD| sin 45◦ + |FAC | sin 70◦ + |FAB | sin 60◦ = |W|,|FAD| cos 45◦ + |FAC | cos 70◦ − |FAB | cos 60◦ = 0.
Setting |W| = 2000 lb and |FAC | = 600 lb and solving, we obtain FAB =1202 lb, FAD = 559 lb.
Full file at https://fratstock.euProblem 2.13 The wires in Problem 2.12 will safelysupport the weight if the magnitude of the vector compo-nent of F parallel to each wire does not exceed 2000 lb.Based on this criterion, how large can the magnitude ofF be? What are the corresponding magnitudes of thevector components of F parallel to the three wires?
Solution: From Problem 2.12 above, we have FAC = 600 lb., FAB =1202 lb. The largest force is FAB = 1202 lb. We want this value to be 2000 lband to have all other values scaled accordingly. Hence, we multiply all forcesby k = 2000
1202 = 1.664. Multiplying all of the forces in Problem 2.12 by thisfactor, we get
FAB = 2000 lb, FAC = 999 lb,
FAD = 931 lb, and F = 3329 lb.
Problem 2.14 Two vectors rA and rB have magni-tudes |rA| = 30 m and |rB | = 40 m. Determine themagnitude of their sum rA + rB
(a) if rA and rB have the same direction.(b) if rA and rB are perpendicular.
Solution: The vector constructions are shown.(a) The magnitude of the sum is the sum of the magnitudes for vectors in thesame direction:
|rA + rB | = 30 + 40 = 70 m
From the cosine law (which reduces to the Pythagorean Theorem for a righttriangle)
|rA + rB |2 = |r2A + |r2
B = (30)2 + (40)2 = 2500 |rA + rB | = 50 m .
rA
rA
rB
rB (b)
(a)
rA + rB
Problem 2.15 A spherical storage tank is sup-ported by cables. The tank is subjected to threeforces: the forces FA and FB exerted by the ca-bles and the weight W. The weight of the tank is|W| = 600 lb. The vector sum of the forces acting onthe tank equals zero. Determine the magnitudes of FA
and FB (a) graphically and (b) by using trigonometry.
40°
FA
W
FB
20° 20°
Solution: The vector construction is shown.(a) The graphical solution is obtained from the construction by the recognitionthat since the opposite interior angles of the triangle are equal, the sides (magni-tudes of the forces exerted by the cables) are equal. A measurement determinesthe magnitudes. (b) The trigonometric solution is obtained from the law of sines:
|W|sin 140◦ =
|FA|sin 20◦ =
|FB |sin 20◦
Solving:
|FA| = |FB | = |W|(
sin 20sin 140
)= 319.25 . . . = 319.3 lb
40°FA
W
FB
20° 20°
FB
FA
W
20
2020
20
140
Full file at https://fratstock.euProblem 2.16 The rope ABC exerts the forces FBA
and FBC on the block at B. Their magnitudes are|FBA| = |FBC | = 800 N. Determine |FBA + FBC |(a) graphically and (b) by using trigonometry.
20°
FBC
FBA
B
C
A
B
Solution: The vector graphical construction is shown.(a) The angles are derived from the rule that for equal legs in a tri-
angle the opposite interior angles are equal, and the rule that thesum of the interior angles is 180 deg. Thus from the problemstatement the 70◦ angle is determined; from the equality of an-gles and the sum of interior angles the other two 55◦ angles arederived. The magnitude of the sum of the two vectors is thenmeasured from the graph.
(b) The trigonometric solution follows from the law of sines:
|FAB + FBC |sin 70◦ =
|FAB |sin 55◦ =
|FBC |sin 55◦ .
Solve:
|FAB + FBC | = |FAB |(
sin 70◦
sin 55◦
)= 800(1.1471 . . .)
= 917.72 . . . |FA + FB | = 917.7 N
20°
20°
55°
70°
55°
20°
FBC
FBA
FBA + FBC
B
Full file at https://fratstock.euProblem 2.17 Two snowcats tow a housing unit to anew location at McMurdo Base, Antarctica. (The topview is shown. The cables are horizontal.) The sum ofthe forces FA and FB exerted on the unit is parallel tothe line L, and |FA| = 1000 lb. Determine |FB | and|FA+FB | (a) graphically and (b) by using trigonometry.
50° 30°FA
FB
L
TOP VIEW
Solution: The graphical construction is shown. The sum ofthe interior angles must be 180◦. (a) The magnitudes of |FB | and|FA + FB | are determined from measurements. (b) The trigonomet-ric solution is obtained from the law of sines:
|FA + FB |sin 100
=|FA|sin 30
=|FB |sin 50
from which |FB | = |FA|(
sin 50sin 30
)= 1000(1.532) = 1532 lb
|FA + FB | = |FA|(
sin 100sin 30
)= 1000(1.9696) = 1970 lb
50° 30°FA
FB
L
TOP VIEW
FB
FA
FB
FA + FB
38°38°
50°50°
156°38°
Problem 2.18 A surveyor determines that the horizon-tal distance from A to B is 400 m and that the horizontaldistance from A to C is 600 m. Determine the magnitudeof the horizontal vector rBC from B to C and the angleα (a) graphically and (b) by using trigonometry.
East
North
60°
20°
α
C
B
A
rBC
Solution: (a) The graphical solution is obtained by drawing thefigure to scale and measuring the unknowns. (b) The trigonometricsolution is obtained by breaking the figure into three separate righttriangles. The magnitude |rBC | is obtained by the cosine law:
|rBC |2 = (400)2 + (600)2 − 2(400)(600) cos 40◦
or |rBC | = 390.25 = 390.3 m
The three right triangles are shown. The distance BD is BD =(400) sin 60◦ = 346.41 m. The distance CE is CE =600 sin 20◦ = 205.2 m. The distance FC is FC = (346.4 −205.2) = 141.2 m.
The angle α is sin α = 141.2390.3 = 0.36177 . . ., or α = 21.2◦
60°20°
C
B F
A D E
Full file at https://fratstock.euProblem 2.19 The vector r extends from point A tothe midpoint between points B and C. Prove that
r =12(rAB + rAC).
A
C
B
rAC
r
rAB
Solution: The proof is straightforward:
r = rAB + rBM , and r = rAC + rCM .
Add the two equations and note that rBM + rCM = 0, since the twovectors are equal and opposite in direction.
Thus 2r = rAC + rAB , or r =(
12
)(rAC + rAB)
rAC
rAB
rM
C
A
Problem 2.20 By drawing sketches of the vectors,explain why
U + (V + W) = (U + V) + W.
Solution: Additive associativity for vectors is usually given as anaxiom in the theory of vector algebra, and of course axioms are notsubject to proof. However we can by sketches show that associativityfor vector addition is intuitively reasonable: Given the three vectors tobe added, (a) shows the addition first of V+W, and then the additionof U. The result is the vector U + (V + W).(b) shows the addition of U+V, and then the addition of W, leadingto the result (U + V) + W.The final vector in the two sketches is the same vector, illustrating thatassociativity of vector addition is intuitively reasonable.
(a)
U
W
V
U
W
V
V+W
U+V
U+[V+W]
[U+V]+W
(b)
Problem 2.21 A force F = 40 i − 20 j (N). What isits magnitude |F|?Strategy: The magnitude of a vector in terms of itscomponents is given by Eq. (2.8).
Solution: |F| =√
402 + 202 = 44.7 N
Problem 2.22 An engineer estimating the componentsof a force F = Fx i + Fy j acting on a bridge abutmenthas determined that Fx = 130 MN, |F| = 165 MN, andFy is negative. What is Fy?
Solution:
|F| =√
|Fx|2 + |Fy |2
Thus |Fy | =√
|F|2 − |Fx|2 (mN)
|Fy | =√
1652 − 1302 (mN)
|Fy | = 101.6 mN
Fy = −102 mN
Full file at https://fratstock.euProblem 2.23 A support is subjected to a force F =Fx i + 80j (N). If the support will safely support a forceof 100 N, what is the allowable range of values of thecomponent Fx?
Solution: Use the definition of magnitude in Eq. (2.8) and reduce alge-braically.100 ≥ √
(Fx)2 + (80)2, from which (100)2 − (80)2 ≥ (Fx)2.
Thus |Fx| ≤ √3600, or −60 ≤ (Fx) ≤ +60 (N)
Problem 2.24 If FA = 600i − 800j (kip) and FB =200i − 200j (kip), what is the magnitude of the forceF = FA − 2FB?
Solution: Take the scalar multiple of FB , add the components of the twoforces as in Eq. (2.9), and use the definition of the magnitude. F = (600 −2(200))i + (−800 − 2(−200))j = 200i − 400j
|F| =√
(200)2 + (−400)2 = 447.2 kip
Problem 2.25 If FA = i − 4.5j (kN) and FB =−2i − 2j (kN), what is the magnitude of the force F =6FA + 4FB?
Solution: Take the scalar multiples and add the components.
F = (6 + 4(−2))i + (6(−4.5) + 4(−2))j = −2i − 35j, and
|F| =√
(−2)2 + (−35)2 = 35.1 kN
Problem 2.26 Two perpendicular vectors U and V liein the x-y plane. The vector U = 6i− 8j and |V| = 20.What are the components of V?
Solution: The two possible values of V are shown in the sketch.The strategy is to (a) determine the unit vector associated with U,(b) express this vector in terms of an angle, (c) add ±90◦ to thisangle, (d) determine the two unit vectors perpendicular to U, and(e) calculate the components of the two possible values of V. The unitvector parallel to U is
eU =6i√
62 + (−8)2− 8j√
62 + (−8)2= 0.6i − 0.8j
Expressed in terms of an angle,
eU = i cos α − j sin α = i cos(53.1◦) − j sin(53.1◦)
Add ±90◦ to find the two unit vectors that are perpendicular to thisunit vector:
ep1 = i cos(143.1◦) − j sin(143.1◦) = −0.8i − 0.6j
ep2 = i cos(−36.9◦) − j sin(−36.9◦) = 0.8i + 0.6j
Take the scalar multiple of these unit vectors to find the two vectorsperpendicular to U.
V1 = |V|(−0.8i − 0.6j) = −16i − 12j.
The components are Vx = −16, Vy = −12
V2 = |V|(0.8i + 0.6j) = 16i + 12j.
The components are Vx = 16, Vy = 12
y
x6
8U
V2
V1
Full file at https://fratstock.euProblem 2.27 A fish exerts a 40-N force on the linethat is represented by the vector F. Express F in termsof components using the coordinate system shown.
y
x
60°
F
Solution:
Fx = |F| cos 60◦ = (40)(0.5) = 20 (N)
Fy = −|F | sin 60◦ = −(40)(0.866) = −34.6 (N)
F = 20i − 34.6j (N)
y
x
60°
F
Problem 2.28 A person exerts a 60-lb force F to pusha crate onto a truck. Express F in terms of components.
F 20°
x
y
Solution: The strategy is to express the force F in terms of the angle. Thus
F = (i|F| cos(20◦) + j|F| sin(20◦))
F = (60)(0.9397i + 0.342j) or F = 56.4i + 20.5j (lb)
F 20°
x
y
Problem 2.29 The missile’s engine exerts a 260-kNforce F. Express F in terms of components using thecoordinate system shown.
y
x
F
40°
Solution:
Fx = |F| cos 40◦
Fx = 199 N
Fy = |F| sin 40◦
Fy = 167 N
F = 199i + 167j (N) y
F
40°
x
Full file at https://fratstock.euProblem 2.30 The coordinates of two points A and Bof a truss are shown. Express the position vector frompoint A to point B in terms of components.
A
B
y
x(2, 1) m
(6, 4) m
Solution: The strategy is find the distance along each axis bytaking the difference between the coordinates.
rAB = (2 − 6)i + (1 − 4)j = −4i − 3j (m)A(6, 4)
B(2, 1)
Problem 2.31 The points A, B, . . . are the joints of thehexagonal structural element. Let rAB be the positionvector from joint A to joint B, rAC the position vectorfrom joint A to joint C, and so forth. Determine thecomponents of the vectors rAC and rAF .
2 m
x
y
A B
E
F C
D
Solution: Use the xy coordinate system shown and find the loca-tions of C and F in those coordinates. The coordinates of the pointsin this system are the scalar components of the vectors rAC and rAF .For rAC , we have
rAC = rAB + rBC = (xB − xA)i + (yB − yA)j
+ (xC − xB)i + (yC − yB)j
or rAC = (2m − 0)i + (0 − 0)j + (2m cos 60◦ − 0)i
+ (2m cos 60◦ − 0)j,
giving
rAC = (2m + 2m cos 60◦)i + (2m sin 60◦)j. For rAF , we have
rAF = (xF − xA)i + (yF − yA)j
= (−2m cos 60◦xF − 0)i + (2m sin 60◦ − 0)j.
y
xBA
FC
2 m
rACrAF
E D
Full file at https://fratstock.euProblem 2.32 For the hexagonal structural element inProblem 2.31, determine the components of the vectorrAB − rBC .
Solution: rAB − rBC .The angle between BC and the x-axis is 60◦.
2 m
x
y
A B
E
F C
D
rBC = 2 cos(60◦)i + 2(sin(60◦)j (m)
rBC = 1i + 1.73j (m)
rAB − rBC = 2i − 1i − 1.73j (m)
rAB − rBC = 1i − 1.73j (m)
Problem 2.33 The coordinates of point A are (1.8,3.0) m. The y coordinate of point B is 0.6 m and themagnitude of the vector rAB is 3.0 m. What are thecomponents of rAB?
x
y
A
B
rAB
Solution: Let the x-component of point B be xB . The vectorfrom A to B can be written as
rAB = (xB − xA)i + (yB − yA)j (m)
or rAB = (xB − 1.8)i + (0.6 − 3.0)j (m)
rAB = (xB − 1.8)i − 2.4j (m)
We also know |rAB | = 3.0 m. Thus
32 = (xB − 1.80)2 + (−2.4)2
Solving, xB = 3.60. Thus
rAB = 1.80i − 2.40j (m)
Full file at https://fratstock.euProblem 2.34 (a) Express the position vector frompoint A of the front-end loader to point B in terms ofcomponents.(b) Express the position vector from point B to point Cin terms of components.(c) Use the results of (a) and (b) to determine the distancefrom point A to point C.
45 in.98 in.
50 in.55 in.35 in.
A
50 in.
y
x
CB
Solution: The coordinates are A(50, 35); B(98, 50); C(45, 55).(a) The vector from point A to B:
rAB = (98 − 50)i + (50 − 35)j = 48i + 15j (in.)
(b) The vector from point B to C is
rBC = (45 − 98)i + (55 − 50)j = −53i + 5j (in.).
(c) The distance from A to C is the magnitude of the sum of thevectors,
rAC = rAB + rBC = (48 − 53)i + (15 + 5)j = −5i + 20j.
The distance from A to C is
|rAC | =√
(−5)2 + (20)2 = 20.62 in.
45 in.
98 in.
50 in.
55 in.
35 in.
A
50 in.
y
x
C B
Problem 2.35 Consider the front-end loader in Prob-lem 2.34. To raise the bucket, the operator increases thelength of the hydraulic cylinder AB. The distance be-tween points B and C remains constant. If the lengthof the cylinder AB is 65 in., what is the position vectorfrom point A to point B?
Solution: Assume that the two points A and C are fixed. Thestrategy is to determine the unknown angle θ from the geometry. FromProblem 2.34 |rAC | = 20.6 and the angle β is tan β =
( −205
)=
−4, β = 76◦. Similarly, |rCB | =√
532 + 52 = 53.2. The angle ais found from the cosine law:
cos α =(20.6)2 + (65)2 − (53.2)2
2(20.6)(65)= 0.6776,
α = 47.3◦. Thus the angle θ is
θ = 180◦ − 47.34◦ − 75.96◦ = 56.69 . . . = 56.7◦. The vector
rAB = 65(i cos θ + j sin θ) = 35.69 . . . i + 54.32 . . . j
= 35.7i + 54.3j (in.)
A
B
C
53.2
6520.6 αθ
β
Full file at https://fratstock.euProblem 2.36 Determine the position vector rAB interms of its components if: (a) θ = 30◦, (b) θ = 225◦.
60 mm 150 mm
x
y
A
C
BrAB
rBC
Solution:
(a) rAB = (60) cos(30◦)i + (60) sin(30◦)j, or
rAB = 51.96i + 30j mm. And
(b) rAB = (60) cos(225◦)i + (60) sin(225◦)j or
rAB = −42.4i − 42.4j mm. FBCFAB
60mm
150mm
y
xC F
A
B
β
Problem 2.37 In problem 2.36 determine the positionvector rBC in terms of its components if: (a) θ = 30◦,(b) θ = 225◦.
Solution:(a) From Problem 2.36, rAB = 51.96i + 30j mm. Thus, the
coordinates of point B are (51.96, 30) mm. The vector rBC isgiven by rBC = (xC − xB)i + (yC − yB)j, whereyC = 0.The magnitude of the vector rBC is 150 mm. Using these facts,we find that yBC = −30 mm, and xBC = 146.97 mm.
(b) rAB = (60) cos(225◦)i + (60) sin(225◦)j or
rAB = −42.4i − 42.4j mm.
From Problem 2.36, rAB = −42.4i − 42.4j mm. Thus, thecoordinates of point B are (−42.4, −42.4) mm. The vector rBC
is given by rBC = (xC −xB)i+(yC −yB)j, where yC = 0.The magnitude of the vector rBC is 150 mm. Using these facts,we find that yBC = 42.4 mm, and xBC = 143.9 mm.
Full file at https://fratstock.euProblem 2.38 A surveyor measures the location ofpoint A and determines that rOA = 400i + 800j (m).He wants to determine the location of a point B so that|rAB | = 400 m and |rOA + rAB | = 1200 m. What arethe cartesian coordinates of point B?
x
y
A
B
O
rAB
rOA
Proposedroadway
N
Solution: Two possibilities are: The point B lies west of point A,or point B lies east of point A, as shown. The strategy is to determinethe unknown angles α, β, and θ. The magnitude of OA is
|rOA| =√
(400)2 + (800)2 = 894.4.
The angle β is determined by
tan β =800400
= 2, β = 63.4◦.
The angle α is determined from the cosine law:
cos α =(894.4)2 + (1200)2 − (400)2
2(894.4)(1200)= 0.9689.
α = 14.3◦. The angle θ is θ = β ± α = 49.12◦, 77.74◦.The two possible sets of coordinates of point B are{
rOB = 1200(i cos 77.7 + j sin 77.7) = 254.67i + 1172.66j (m)rOB = 1200(i cos 49.1 + j sin 49.1) = 785.33i + 907.34j (m)
The two possibilities lead to B(254.7 m, 1172.7 m) or B(785.3 m,907.3 m)
B
y
x0
θα
α
β
BA
Problem 2.39 Bar AB is 8.5 m long and bar AC is6 m long. Determine the components of the positionvector rAB from point A to point B.
A
B Cx
y
3 m
Solution: The key to this solution is to find the coordinates ofpoint A. We know the lengths of all three sides of the triangle. Thelaw of cosines can be used to give us the angle θ.
|rAC |2 = |rAB |2 + |rBC |2 − 2|rAB ||rBC | cos θ
62 = 8.52 + 32 − 2(8.5)(3) cos θ
cos θ = 0.887 θ = 27.5◦
rAB = |rAB | cos θi − |rAB | sin θj
rAB = (8.5) cos 27.5◦i − (8.5) sin 27.5◦j (m)
rAB = 7.54i − 3.92j m
y
x
3 m
B Cθ
A
8.5 m
6 m
Full file at https://fratstock.euProblem 2.40 For the truss in Problem 2.39, determinethe components of a unit vector eAC that points frompoint A toward point C.Strategy: Determine the components of the positionvector from point A to point C and divide the positionvector by its magnitude.
Solution: From the solution of Problem 2.39, point A is locatedat (7.54, −3.92). From the diagram, Point C is located at (3.0, 0).The vector from A to C is
rAC = (xC − xA)i + (yC − yA)j (m)
rAC = (3 − 7.54)i + (0 − (−3.92))j (m)
rAC = −4.54i + 3.92j (m)
and |rAC | = 6 m
Thus eAC =rAC
|rAC | = −4.546
i +3.926
j
eAC = −0.757i + 0.653j
Problem 2.41 The x and y coordinates of points A,B, and C of the sailboat are shown.(a) Determine the components of a unit vector that is
parallel to the forestay AB and points from A to-ward B.
(b) Determine the components of a unit vector that isparallel to the backstay BC and points from C to-ward B.
y
x
B (4,13) m
C(9,1) m
A(0,1.2) m
Solution:
rAB = (xB − xA)i + (yB − yA)j
rCB = (xB − xC)i + (yC − yB)j
Points are: A (0, 1.2), B (4, 13) and C (9, 1)Substituting, we get
rAB = 4i + 11.8j (m), |rAB | = 12.46 (m)
rCB = −5i + 12j (m), |rCB | = 13 (m)
The unit vectors are given by
eAB =rAB
|rAB | and eCB =rCB
|rCB |Substituting, we get
eAB = 0.321i + 0.947j
eCB = −0.385i + 0.923j
y
x
B (4, 13) m
C(9, 1) mA
(0, 1.2) m
Full file at https://fratstock.euProblem 2.42 Consider the force vector F = 3i − 4j(kN). Determine the components of a unit vector e thathas the same direction as F.
Solution: The magnitude of the force vector is
|F| =√
32 + 42 = 5.
The unit vector is
e =F|F| =
35i − 4
5j = 0.6i − 0.8j As a check, the magnitude:
|e| =√
0a.62 + 0.82 = 1
Problem 2.43 Determine the components of a unitvector that is parallel to the hydraulic actuator BC andpoints from B toward C.
1 m
0.6 m Scoop
A B
D
C
0.15 m
0.6 m
1 m
x
y
Solution: Point B is at (0.75, 0) and point C is at (0, 0.6). Thevector
rBC = (xC − xB)i + (yC − yB)j
rBC = (0 − 0.75)i + (0.6 − 0)j (m)
rBC = −0.75i + 0.6j (m)
|rBC | =√
(0.75)2 + (0.6)2 = 0.960 (m)
eBC =rBC
|rBC | =−0.750.96
i +0.60.96
j
eBC = −0.781i + 0.625j
1 m
0.6 m Scoop
A B
D
C
0.15 m
0.6 m
1 m
x
y
Problem 2.44 The hydraulic actuator BC in Problem2.43 exerts a 1.2-kN force F on the joint at C that isparallel to the actuator and points from B toward C.Determine the components of F.
Solution: From the solution to Problem 2.43,
eBC = −0.781i + 0.625j
The vector F is given by F = |F|eBC
F = (1.2)(−0.781i + 0.625j) (k · N)
F = −937i + 750j (N)
Full file at https://fratstock.euProblem 2.45 A surveyor finds that the length of theline OA is 1500 m and the length of line OB is 2000 m.(a) Determine the components of the position vector
from point A to point B.(b) Determine the components of a unit vector that
points from point A toward point B.
x
y
60°
B
A
O
Proposed bridge
River30°
N
Solution: We need to find the coordinates of points A and B
rOA = 1500 cos 60◦i + 1500 sin 60◦j
rOA = 750i + 1299j (m)
Point A is at (750, 1299) (m)
rOB = 2000 cos 30◦i + 2000 sin 30◦j (m)
rOB = 1723i + 1000j (m)
Point B is at (1732, 1000) (m)(a) The vector from A to B is
rAB = (xB − xA)i + (yB − yA)j
rAB = 982i − 299j (m)
(b) The unit vector eAB is
eAB =rAB
|rAB | =982i − 299j
1026.6
eAB = 0.957i − 0.291j
x
y
A
B
60°
30°
O
Proposedbridge
River
N
Full file at https://fratstock.euProblem 2.46 The positions at a given time of theSun (S) and the planets Mercury (M), Venus (V), andEarth (E) are shown. The approximate distance fromthe Sun to Mercury is 57 × 106 km, the distance fromthe Sun to Venus is 108×106 km, and the distance fromthe Sun to the Earth is 150 × 106 km. Assume that theSun and planets lie in the x − y plane. Determine thecomponents of a unit vector that points from the Earthtoward Mercury.
40°
20°
S M
E
V
x
y
Solution: We need to find rE and rM in the coordinates shown
rE = |rE |(− sin 20◦i) + |rE |(cos 20◦)j (km)
rM = |rM | cos 0◦i (km)
rE = (−51.3 × 106)i + (141 × 106)j (km)
rM = 57 × 106i (km)
rEM = (xM − xE)i + (yM − yE)j (km)
rEM = (108.3 × 106)i − (141 × 106j) (km)
|rEM | = 177.8 × 106 (km)
eEM =rEM
|rEM | = +0.609i − 0.793j
40°
20°
S M
E
V
x
y
Problem 2.47 For the positions described in Prob-lem 2.46, determine the components of a unit vector thatpoints from Earth toward Venus.
Solution: From the solution to Problem 2.47,
rE = (−51.3 × 106)i + (141 × 106)j (km)
The position of Venus is
rV = −|rV | cos 40◦i − |rV | sin 40◦j (km)
rV = (−82.7 × 106)i − (69.4 × 106)j (km)
rEV = (xV − xE)i + (yV − yE)j (km)
rEV = (−31.4 × 106)i − (210.4 × 106)j (km)
|rEV | = 212.7 × 106 (km)
eEV =rEV
|rEV |eEV = −0.148i − 0.989j
Full file at https://fratstock.euProblem 2.48 The rope ABC exerts forces FBA
and FBC on the block at B. Their magnitudes are|FBA| = |FBC | = 800 N. Determine the magnitudeof the vector sum of the forces by resolving the forcesinto components, and compare your answer with that ofProblem 2.16.
20°
FBC
FBA
B
C
A
B
Solution: The strategy is to use the magnitudes and the angles todetermine the force vectors, and then to determine the magnitude oftheir sum. The force vectors are:
FBA = 0i − 800j, and
FBC = 800(i cos 20◦ + j sin 20◦) = 751.75i + 273.6j
The sum is given by: FBA + FBC = 751.75i − 526.4j
The magnitude is given by
|FBA + FBC | =√
(751.75)2 + (526.4)2 = 917.7 N
B
FBA
FBC
x
y
20°
Problem 2.49 The magnitudes of the forces are |F1| =|F2| = |F3| = 5 kN. What is the magnitude of the vectorsum of the three forces?
F1
F2
F3
30°
45°
x
y
Solution: The strategy is to use the magnitudes and the angles todetermine the force vectors, and then to take the magnitude of theirsum. The force vectors are:
F1 = 5i + 0j (kN),
F2 = 5(i cos(−45◦) + j sin(−45◦)) = 3.54i − 3.54j
F3 = 5(i cos 210◦ + j sin 210◦) = −4.33i − 2.50j
The sum is given by F1 + F2 + F3 = 4.21i − 6.04j and the mag-nitude is
|F1 + F2 + F3| =√
(4.21)2 + (6.04)2 = 7.36 kN
y
xF2
F1
F3
30°
45°
Full file at https://fratstock.euProblem 2.50 Four groups engage in a tug-of-war.The magnitudes of the forces exerted by groups B, C,and D are |FB | = 800 lb, |FC | = 1000 lb, |FD| =900 lb. If the vector sum of the four forces equals zero,what are the magnitude of FA and the angle α?
FB
x
y
70°30°
20°α
FC
FA
FD
Solution: The strategy is to use the angles and magnitudes todetermine the force vector components, to solve for the unknown forceFA and then take its magnitude. The force vectors are
FB = 800(i cos 110◦ + j sin 110◦) = −273.6i + 751.75j
FC = 1000(i cos 30◦ + j sin 30◦) = 866i + 500j
FD = 900(i cos(−20◦) + j sin(−20◦)) = 845.72i − 307.8j
FA = |FA|(i cos(180 + α) + j sin(180 + α))
= |FA|(−i cos α − j sin α)
The sum vanishes:
FA + FB + FC + FD = i(1438.1 − |FA| cos α)
+ j(944 − |FA| sin α) = 0
From which FA = 1438.1i + 944j. The magnitude is
|FA| =√
(1438)2 + (944)2 = 1720 lb
The angle is: tan α = 9441438 = 0.6565, or α = 33.3◦
y
x
FB
FA
FD
FC
30°
20°
70°
α
Problem 2.51 The total thrust exerted on the launchvehicle by its main engines is 200,000 lb parallel to they axis. Each of the two small vernier engines exert athrust of 5000 lb in the directions shown. Determine themagnitude and direction of the total force exerted on thebooster by the main and vernier engines.
30°
y
15°
x
Vernierengines
Solution: The strategy is to use the magnitudes and angles todetermine the force vectors. The force vectors are:
FME = 0i − 200j (kip)
FLV = 5(i cos 240◦ + j sin 240◦) = −2.5i − 4.33j (kip)
FRV = 5(i cos 285◦ + j sin 285◦) = 1.29i − 4.83j (kip)
The sum of the forces:
FME + FRV + FLV = −1.21i − 209.2j (kip).
The magnitude of the sum is
|FR| =√
(1.21)2 + (209.2)2 = 209.2 (kip)
The direction relative to the y-axis is
tan α =1.21209.2
= 0.005784, or α = 0.3314◦
measured clockwise from the negative y-axis.
30° 15°
5k 5k
200k
Full file at https://fratstock.euProblem 2.52 The magnitudes of the forces acting onthe bracket are |F1| = |F2| = 2 kN. If |F1 + F2| =3.8 kN, what is the angle α? (Assume 0 ≤ α ≤ 90◦)
F2
F1
α
Solution:
Let |F1| = |F2| = a = 2 kn
and |F1 + F2| = b
F2
F1
α
α
β αF1
F1 + F2
F2
Angle β is given by
α
2+ β +
α
2= 180◦
β = 180◦ − α b = 3.8 kN
a = 2 kN
From the law of cosines
b2 = a2 + a2 − 2a2 cos β
β = 143.6◦
α = 180◦ − β α = 36.4◦
Full file at https://fratstock.euProblem 2.53 The figure shows three forces acting ona joint of a structure. The magnitude of Fc is 60 kN, andFA + FB + FC = 0. What are the magnitudes of FA
and FB?
y
x
FB
FC
FA
15°
40°
Solution: We need to write each force in terms of its components.
FA = |FA| cos 40i + |FA| sin 40j (kN)
FB = |FB | cos 195◦i + |FB | sin 195j (kN)
FC = |FC | cos 270◦i + |FC | sin 270◦j (kN)
Thus FC = −60j kNSince FA +FB +FC = 0, their components in each direction mustalso sum to zero.{
FAx + FBx + FCx = 0FAy + FBy + FCy = 0
Thus,{ |FA| cos 40◦ + |FB | cos 195◦ + 0 = 0|FA| sin 40◦ + |FB | sin 195◦ − 60 (kN) = 0
Solving for |FA| and |FB |, we get
|FA| = 137 kN, |FB | = 109 kN
FB
FC
FA195°
270°
40°x
y
x
FB
FC
FA
15°
40°
Problem 2.54 Four forces act on a beam. The vectorsum of the forces is zero. The magnitudes |FB | = 10 kNand |FC | = 5 kN. Determine the magnitudes of FA
and FD.
FD
30°
FB FCFA
Solution: Use the angles and magnitudes to determine the vectors,and then solve for the unknowns. The vectors are:
FA = |FA|(i cos 30◦ + j sin 30◦) = 0.866|FA|i + 0.5|FA|jFB = 0i − 10j, FC = 0i + 5j, FD = −|FD|i + 0j.
Take the sum of each component in the x- and y-directions:∑Fx = (0.866|FA| − |FD|)i = 0
and∑
Fy = (0.5|FA| − (10 − 5))j = 0.
From the second equation we get |FA| = 10 kN . Using this value in the first
equation, we get |FD| = 8.7 kN
FD
30°νFB FCFA
Full file at https://fratstock.euProblem 2.55 Six forces act on a beam that forms partof a building’s frame. The vector sum of the forces iszero. The magnitudes |FB | = |FE | = 20 kN, |FC | =16 kN, and |FD| = 9 kN. Determine the magnitudes ofFA and FG.
70°40°40° 50°
FA FC FDFG
FEFB
Solution: Write each force in terms of its magnitude and directionas
F = |F| cos θi + |F| sin θj
where θ is measured counterclockwise from the +x-axis.Thus, (all forces in kN)
FA = |FA| cos 110◦i + |FA| sin 110◦j (kN)
FB = 20 cos 270◦i + 20 sin 270◦j (kN)
FC = 16 cos 140◦i + 16 sin 140◦j (kN)
FD = 9 cos 40◦i + 9 sin 40◦j (kN)
FE = 20 cos 270◦i + 20 sin 270◦j (kN)
FG = |FG| cos 50◦i + |FG| sin 50◦j (kN)
We know that the x components and y components of the forces mustadd separately to zero.Thus{
FAx + FBx + FCx + FDx + FEx + FGx = 0FAy + FBy + FCy + FDy + FEy + FGy = 0
{ |FA| cos 110◦ + 0 − 12.26 + 6.89 + 0 + |FG| cos 50◦ = 0|FA| sin 110◦ − 20 + 10.28 + 5.79 − 20 + |FG| sin 50◦ = 0
Solving, we get
|FA| = 13.0 kN |FG| = 15.3 kN
x
y
θ
70°40°40° 50°
FA FC FDFG
FEFB
Full file at https://fratstock.euProblem 2.56 The total weight of the man and parasailis |W| = 230 lb. The drag force D is perpendicular tothe lift force L. If the vector sum of the three forces iszero, what are the magnitudes of L and D?
20°
x
y
L
D
W
Solution: Three forces in equilibrium form a closed triangle. In this in-stance it is a right triangle. The law of sines is
|W|sin 90◦ =
|L|sin 70◦ =
|D|sin 20◦
From which:
|L| = |W| sin 70◦ = (230)(0.9397) = 216.1 lb
|D| = |W| sin 20◦ = (230)(0.3420) = 78.66 lb
x
y
W
L
D
20°
W
L
D
20°
Problem 2.57 Two cables AB and CD extend fromthe rocket gantry to the ground. Cable AB exerts a forceof magnitude 10,000 lb on the gantry, and cable CDexerts a force of magnitude 5000 lb.(a) Using the coordinate system shown, express each of
the two forces exerted on the gantry by the cables interms of scalar components.
(b) What is the magnitude of the total force exerted onthe gantry by the two cables?
x
y
A
C
D B
40°
30°
Solution: Use the angles and magnitudes to determine the com-ponents of the two forces, and then determine the magnitude of theirsum. The forces:
(a) FAB = 10(i cos(−50◦) + j sin(−50◦))
= 6.428i − 7.660j kip
FCD = 5(i cos(−60◦) + j sin(−60◦)) = 2.50i − 4.330j kip
The sum: FAB + FCD = 8.928i − 11.990j,
The magnitude is:
(b) |FAB + FCD| =√
(8.928)2 + (11.99)2 = 14.95 kip
A
C
D B
40°50 °
60°
30°
Full file at https://fratstock.euProblem 2.58 The cables A, B, and C help support apillar that forms part of the supports of a structure. Themagnitudes of the forces exerted by the cables are equal:|FA| = |FB | = |FC |. The magnitude of the vector sumof the three forces is 200 kN. What is |FA|? FA
FB
FC
4 m
AB
C6 m
4 m 4 m
Solution: Use the angles and magnitudes to determine the vectorcomponents, take the sum, and solve for the unknown. The anglesbetween each cable and the pillar are:
θA = tan−1(
4 m
6 m
)= 33.7◦,
θB = tan−1(
86
)= 53.1◦
θC = tan−1(
126
)= 63.4◦.
Measure the angles counterclockwise form the x-axis. The force vec-tors acting along the cables are:
FA = |FA|(i cos 303.7◦ + j sin 303.7◦) = 0.5548|FA|i − 0.8319|FA|jFB = |FB |(i cos 323.1◦ + j sin 323.1◦) = 0.7997|FB |i − 0.6004|FB |jFC = |FC |(i cos 333.4◦ + j sin 333.4◦) = 0.8944|FC |i−0.4472|FC |jThe sum of the forces are, noting that each is equal in magnitude, is∑
F = (2.2489|FA|i − 1.8795|FA|j).The magnitude of the sum is given by the problem:
200 = |FA|√
(2.2489)2 + (1.8795)2 = 2.931|FA|,from which |FA| = 68.24 kN
4 m
A B C6 m
4 m 4 m
Full file at https://fratstock.euProblem 2.59 The cable from B to A on the sailboatshown in Problem 2.41 exerts a 230-N force at B. Thecable from B to C exerts a 660-N force at B. What isthe magnitude of the total force exerted at B by the twocables? What is the magnitude of the downward force(parallel to the y axis) exerted by the two cables on theboat’s mast?
Solution: Find unit vectors in the directions of the two forces-express the forces in terms of magnitudes times unit vectors-add theforces.Unit vectors:
eBA =rBA
|rBA| =(xA − xB)i + (yA − yB)j√(xA − xB)2 + (yA − yB)2
=(0 − 4)i + (1.2 − 13)j√
42 + 11.82
eBA = −0.321i − 0.947j
Similarly,
eBC = 0.385i − 0.923j
FBA = |FBA|eBA = −73.8i − 217.8j kN
FBC = |FBC |eBC = 254.1i − 609.2j kN
Adding
F = FBA + FBC = 180.3i − 827j kN
|F| =√
F 2x + F 2
y = 846 kN (Total force)
Fy = −827 kN (downward force)
y
x
B (4, 13) m
C(9, 1) mA
(0, 1.2) m
Problem 2.60 The structure shown forms part of atruss designed by an architectural engineer to supportthe roof of an orchestra shell. The members AB, AC,and AD exert forces FAB , FAC , and FAD on the jointA. The magnitude |FAB | = 4 kN. If the vector sum ofthe three forces equals zero, what are the magnitudes ofFAC and FAD?
FAB
A
(– 4, 1) mB
C
D
x
y
FAC
FAD
(–2, –3) m
(4, 2) m
Solution: Determine the unit vectors parallel to each force:
eAD =−2√
22 + 32i +
−3√22 + 32
j = −0.5547i − 0.8320j
eAC =−4√
42 + 12i +
1√42 + 12
j = −0.9701i + 0.2425j
eAB =4√
42 + 22i +
2√42 + 22
j = 0.89443i + 0.4472j
The forces are FAD = |FAD|eAD, FAC = |FAC |eAC ,FAB = |FAB |eAB = 3.578i + 1.789j. Since the vector sum ofthe forces vanishes, the x- and y-components vanish separately:∑
Fx = (−0.5547|FAD| − 0.9701|FAC | + 3.578)i = 0, and∑
Fy = (−0.8320|FAD| + 0.2425|FAC | + 1.789)j = 0
These simultaneous equations in two unknowns can be solved by anystandard procedure. An HP-28S hand held calculator was used here:
The results: |FAC | = 2.108 kN , |FAD| = 2.764 kN
A
B
C
D
Full file at https://fratstock.euProblem 2.61 The distance s = 45 in.(a) Determine the unit vector eBA that points from B
toward A.(b) Use the unit vector you obtained in (a) to determine
the coordinates of the collar C.
y
x
s
A
B
C
(14, 45) in
(75, 12) in
Solution: The unit vector from B to A is the vector from B to A dividedby its magnitude. The vector from B to A is given by
rBA = (xA − xB)i + (yA − yB)j or rBA = (14 − 75)i + (45 − 12)j in.
Hence, vector from B to A is given by rBA = (−61)i + (33)j in. Themagnitude of the vector from B to A is 69.4 in. and the unit vector from Btoward A is eBA = −0.880i + 0.476j.
s
A
B
C
(14, 45) in
(75, 12) in
y
x
Problem 2.62 In Problem 2.61, determine the x and ycoordinates of the collar C as functions of the distance s.
Solution: The coordinates of the point C are given by
xC = xB + s(−0.880) and yC = yB + s(0.476).
Thus, the coordinates of point C are xC = 75 − 0.880s in. and yC =12+0.476s in. Note from the solution of Problem 2.61 above, 0 ≤ s ≤ 69.4 in.
Problem 2.63 The position vector r goes from pointA to a point on the straight line between B and C. Itsmagnitude is |r| = 6 ft. Express r in terms of scalarcomponents.
x
y
A
(7, 9) ft
(12, 3) ft
(3, 5) ft
B
C
r
Solution: Determine the perpendicular vector to the line BCfrom point A, and then use this perpendicular to determine the an-gular orientation of the vector r. The vectors are
rAB = (7 − 3)i + (9 − 5)j = 4i + 4j, |rAB | = 5.6568
rAC = (12 − 3)i + (3 − 5)j = 9i − 2j, |rAC | = 9.2195
rBC = (12 − 7)i + (3 − 9)j = 5i − 6j, |rBC | = 7.8102
The unit vector parallel to BC is
eBC =rBC
|rBC | = 0.6402i − 0.7682j = i cos 50.19◦ − j sin 50.19◦.
Add ±90◦ to the angle to find the two possible perpendicular vectors:
eAP1 = i cos 140.19◦ − j sin 140.19◦, or
eAP2 = i cos 39.8◦ + j sin 39.8◦.
Choose the latter, since it points from A to the line.Given the triangle defined by vertices A, B, C, then the magnitude ofthe perpendicular corresponds to the altitude when the base is the line
BC. The altitude is given by h = 2(area)base . From geometry, the area
of a triangle with known sides is given by
area =√
s(s − |rBC |)(s − |rAC |)(s − |rAB |),
where s is the semiperimeter, s = 12 (|rAC | + |rAB | + |rBC |). Substituting
values, s = 11.343, and area = 22.0 and the magnitude of the perpendicular
is |rAP | = 2(22)7.8102 = 5.6333. The angle between the vector r and the per-
pendicular rAP is β = cos−1 5.63336 = 20.1◦. Thus the angle between the
vector r and the x-axis is α = 39.8 ± 20.1 = 59.1◦ or 19.7◦. The first angleis ruled out because it causes the vector r to lie above the vector rAB , which isat a 45◦ angle relative to the x-axis. Thus:
r = 6(i cos 19.7◦ + j sin 19.7◦) = 5.65i + 2.02j
y
x
B[7,9]
A[3,5]
C[12,3]
P
r
Full file at https://fratstock.euProblem 2.64 Let r be the position vector from pointC to the point that is a distance s meters from point Aalong the straight line between A and B. Express r interms of scalar components. (Your answer will be interms of s.)
x
y
A
(9, 3) m
(10, 9) m
(3, 4) m
B
C
rs
Solution: Determine the ratio of the parts of the line AB and usethis value to determine r. The vectors are:
rAB = (10 − 3)i + (9 − 4)j = 7i + 5j, |rAB | = 8.602
rCA = (3 − 9)i + (4 − 3) = −6i + 1j, |rCA| = 6.0828
rCB = (10 − 9)i + (9 − 3)j = 1i + 6j, |rCB | = 6.0828
The ratio of the magnitudes of the two parts of the line is
|rBP ||rPA| = R =
s
|rBC | − s
Since the ratio is a scalar, then rBP = RrPA, from which (r −rCA) = R(rCB − r).Solve for the vector r, r = RrCB+rCA
1+R. Substitute the values of the
vectors, note that R = s8.602−s
, and reduce algebraically:
r = (0.8138s − 6)i + (0.5813s + 1)j (m) :
Check: An alternate solution: Find the angle of the line AB:
θ = tan−1(
57
)= 35.54◦.
The components of s,
s = |s|(i cos θ + j sin θ) = |s|(0.8138i + 0.5812j).
The coordinates of point P (3+0.8138|s|, 4+0.5812|s|). Subtract coordinatesof point C to get
r = (0.8135|s| − 6)i + (0.5812|s| + 1)j . check .
y
xA[3,4] m
C [9,3] m
B[10,9] m
r8
P
Problem 2.65 A vector U = 3i − 4j − 12k. What isits magnitude?Strategy: The magnitude of a vector is given in termsof its components by Eq. (2.14).
Solution: Use definition given in Eq. (14). The vector magni-tude is
|U| =√
32 + (−4)2 + (−12)2 = 13
Problem 2.66 A force vector F = 20i + 60j − 90k(N). Determine its magnitude.
Solution: Use definition given in Eq. (14). The magnitude of thevector is
|F| =√
(20)2 + (60)2 + (−90)2 = 110 N
Full file at https://fratstock.euProblem 2.67 An engineer determines that an attach-ment point will be subjected to a force F = 20i+Fyj−45k (kN). If the attachment point will safely support aforce of 80-kN magnitude in any direction, what is theacceptable range of values for Fy?
Solution:
802 ≥ F 2x + F 2
y + F 2z
802 ≥ 202 + F 2y + (45)2
To find limits, use equality.
F 2y LIMIT
= 802 − 202 − (45)2
F 2y LIMIT
= 3975
Fy LIMIT = +63.0, −63.0 (kN)
|Fy LIMIT | ≤ 63.0 kN − 63.0 kN ≤ Fy ≤ 63.0 kN
Problem 2.68 A vector U = Uxi + Uyj + Uzk. Itsmagnitude is |U| = 30. Its components are related bythe equations Uy = −2Ux and Uz = 4Uy . Determinethe components.
Solution: Substitute the relations between the components, de-termine the magnitude, and solve for the unknowns. Thus
U = Uxi + (−2Ux)j + (4(−2Ux))k = Ux(1i − 2j − 8k)
where Ux can be factored out since it is a scalar. Take the magnitude,noting that the absolute value of |Ux| must be taken:
30 = |Ux|√
12 + 22 + 82 = |Ux|(8.31).
Solving, we get |Ux| = 3.612, or Ux = ±3.61. The two possiblevectors are
U = +3.61i + (−2(3.61))j + (4(−2)(3.61))k
= 3.61i − 7.22j − 28.9k
U = −3.61i + (−2(−3.61))j
+ 4(−2)(−3.61)k = −3.61i + 7.22j + 28.9k
Problem 2.69 A vector U = 100i + 200j − 600k,and a vector V = −200i+450j+100k. Determine themagnitude of the vector −2U + 3V.
Solution: The resultant is
−2U + 3V = (−2(100) + 3(−200))i + (−2(200) + 3(450))j
+ (−2(−600) + 3(100))k
−2U + 3V = −800i + 950j + 1500k
The magnitude is:
| − 2U + 3V| =√
(−800)2 + (950)2 + (1500)2 = 1947.4
Problem 2.70 Two vectors U = 3i − 2j + 6k andV = 4i + 12j − 3k.(a) Determine the magnitudes of U and V.(b) Determine the magnitude of the vector 3U + 2V.
Solution: The magnitudes:
(a) |U| =√
32 + 22 + 62 = 7 and |V| =√
42 + 122 + 32 = 13The resultant vector
3U + 2V = (9 + 8)i + (−6 + 24)j + (18 − 6)k
= 17i + 18j + 12k
(b) The magnitude |3U + 2V| =√
172 + 182 + 122 = 27.51
Problem 2.71 A vector U = 40i − 70j − 40k.(a) What is its magnitude?(b) What are the angles θx, θy , and θz between U and
the positive coordinate axes?Strategy: Since you know the components of U,you can determine the angles θx, θy , and θz fromEqs. (2.15).
Solution: The magnitude:(a) |U| =
√402 + 702 + 402 = 90
(b) The direction cosines:
U = 90(
4090
i − 7090
j − 4090
)
= 90(0.4444i − 0.7777j − 0.4444k)
U = 90(i cos 63.6◦ + j cos 141.1◦ + k cos 116.4◦)
Full file at https://fratstock.euProblem 2.72 A force F = 600i − 700j + 600k (lb).What are the angles θx, θy , and θz between the vector Fand the positive coordinate axes?
Solution: The magnitude:
|F| =√
6002 + 7002 + 6002 = 1100The unit vector is:
e =F|F| =
6001100
i − 7001100
j +6001100
k = 0.5455i − 0.6364j + 0.5455k
The angles are
θx = cos−1(0.5455) = 56.9◦ , θy = cos−1(−0.6364) = 129.5◦ ,
and θz = cos−1(0.5455) = 56.9◦
Problem 2.73 The cable exerts a 50-lb force F on themetal hook at O. The angle between F and the x axis is40◦, and the angle between F and the y axis is 70◦. Thez component of F is positive.(a) Express F in terms of components.(b) What are the direction cosines of F?
Strategy: Since you are given only two of the an-gles between F and the coordinate axes, you mustfirst determine the third one. Then you can obtainthe components of F from Eqs. (2.15).
y
x
z
O
70
40
F
Solution: Use Eqs. (2.15) and (2.16).The force F = 50(i cos 40◦ + j cos 70◦ + k cos θz). Since12 = cos2 40◦ + cos2 70◦ + cos2 θz , by definition (seeEq. (2.16)) thencos θz = ±√
1 − 0.5868 − 0.1170 = ±0.5442.Thus the components of F are
(a) F = 50(0.7660i + 0.3420j + 0.5442k),
= 38.3i + 17.1j + 27.2k (lb)
(b) the direction cosines are
cos θx = 0.7660, cos θy = 0.3420, cos θz = 0.5442
Problem 2.74 A unit vector has direction cosinescos θx = −0.5 and cos θy = 0.2. Its z component ispositive. Express it in terms of components.
Solution: Use Eq. (2.15) and (2.16). The third direction cosine is
cos θz = ±√
1 − (0.5)2 − (0.2)2 = +0.8426.
The unit vector is
u = −0.5i + 0.2j + 0.8426k
Problem 2.75 The airplane’s engines exert a totalthrust force T of 200-kN magnitude. The angle be-tween T and the x axis is 120◦, and the angle betweenT and the y axis is 130◦. The z component of T ispositive.(a) What is the angle between T and the z axis?(b) Express T in terms of components.
130°
120°
y
x
z
T
y
x
z
Solution: The x- and y-direction cosines are
l = cos 120◦ = −0.5, m = cos 130◦ = −0.6428
from which the z-direction cosine is
n = cosθz = ±√
1 − (0.5)2 − (0.6428)2 = +0.5804.
Thus the angle between T and the z-axis is
(a) θz = cos−1(0.5804) = 54.5◦ , and the thrust is
T = 200(−0.5i − 0.6428j + 0.5804k), or:
(b) T = −100i − 128.6j + 116.1k (kN)
Full file at https://fratstock.euProblem 2.76 The position vector from a point A toa point B is 3i + 4j − 4k (ft). The position vector frompoint A to point C is −3i + 13j − 2k. ft(a) What is the distance from point B to point C?(b) What are the direction cosines of the position vector
from point B to point C?
Solution: The vector from point B to point C is rBC = rAC − rAB .Thus
rBC = (−3 − 3)i + (13 − 4)j + (−2 − (−4))k = −6i + 9j + 2k.
The distance between points B and C is(a) |rBC | =
√62 + 92 + 22 = 11 (ft). The direction cosines are
(b)cos θx = −6
11 = −0.5454, cos θy = 911 = 0.8182,
cos θz = 211 = 0.1818
Problem 2.77 A vector U = 3i − 2j + 6k. Deter-mine the components of the unit vector that has the samedirection as U.
Solution: By definition, the unit vector is the vector whose componentsare the direction cosines of U. (See discussion following Eq. (2.15)). Themagnitude is |U| =
√32 + 22 + 62 = 7. Thus the unit vector is
u = U|U| = 3
7 i − 27 j + 6
7k
Problem 2.78 A force vector F = 3i − 4j − 2k (N).(a) What is the magnitude of F?(b) Determine the components of the unit vector that
has the same direction as F.
Solution: By definition, the unit vector is the vector whose components arethe direction cosines of F. The magnitude is
(a) |F| =√
32 + 42 + 22 = 5.385 (N) The unit vector is
(b) e =3
5.385i − 4
5.385j − 2
5.385k
= 0.5571i − 0.7428j − 0.3714k
Problem 2.79 A force vector F points in the samedirection as the unit vector e = 2
7 i − 67 j − 3
7k. Themagnitude of F is 700 lb. Express F in terms of com-ponents.
Solution: By definition, F = |F|e, where e is a unit vector in the directionof F. (See discussion following Eq. (2.16).) Thus
F = 700( 2
7 i − 67 j − 3
7k)
= 200i − 600j − 300k
Problem 2.80 A force vector F points in the samedirection as the position vector r = 4i + 4j − 7k (m).The magnitude of F is 90 kN. Express F in terms ofcomponents.
Solution: By definition, F = |F|e, where e is a unit vector in the directionof F. Find the unit vector from the position vector. The magnitude is |r| =√
42 + 42 + 72 = 9; the unit vector is e = 49 i + 4
9 j − 79k. The components
are
F = 90(
49i +
49j − 7
9k)
= 40i + 40j − 70k (kN)
Full file at https://fratstock.euProblem 2.81 Astronauts on the space shuttle useradar to determine the magnitudes and direction cosinesof the position vectors of two satellites A and B. Thevector rA from the shuttle to satellite A has magnitude2 km, and direction cosines cos θx = 0.768, cos θy =0.384, cos θz = 0.512. The vector rB from the shuttleto satellite B has magnitude 4 km and direction cosinescos θx = 0.743, cos θy = 0.557, cos θz = −0.371.What is the distance between the satellites?
x
rB
z
B
A rA
y
Solution: The two position vectors are:
rA = 2(0.768i+ 0.384j+ 0.512k)=1.536i + 0.768j + 1.024k (km)
rB = 4(0.743i+ 0.557j− 0.371k)=2.972i + 2.228j − 1.484k (km)
The distance is the magnitude of the difference:
|rA − rB |
=√
(1.536−2.927)2 + (0.768−2.228)2 + (1.024−(−1.484))2
= 3.24 (km)
x
rB
B
A rAy
z
Problem 2.82 Archaeologists measure a pre-Colum-bian ceremonial structure and obtain the dimensionsshown. Determine (a) the magnitude and (b) the di-rection cosines of the position vector from point A topoint B.
4 m
y
10 m
z
b
x
A
C
10 m
B
4 m
8 m
8 m
Solution: The coordinates are A(0, 16, 14), and B(10, 8, 4). Thevector from A to B is
rAB = (10 − 0)i + (8 − 16)j + (4 − 14)k = 10i − 8j − 10k.
The magnitude is
(a) |rAB | =√
102 + 82 + 102 = 16.2 m , and
(b) The direction cosines are
cos θx = 1016.2 = 0.6155,
cos θy = −816.2 = −0.4938,
and cos θz = −1016.2 = −0.6155 .
4 my
10 m
b
x
A
C
10 m
B
4 m
8 m
8 mz
Full file at https://fratstock.euProblem 2.83 Consider the structure described in Pro-blem 2.82. After returning to the United States, an arc-haeologist discovers that he lost the notes containing thedimension b, but other notes indicate that the distancefrom point B to point C is 16.4 m. What are the directioncosines of the vector from B to C?
Solution: The coordinates of B and C are B(10, 8, 4) and C(10 +b, 0,18). The vector from B to C is
rBC = (10 + b − 10)i + (0 − 8)j + (18 − 4)k = bi − 8j + 14k.
The magnitude of this vector is known:
16.4 =√
b2 + 82 + 142 =√
b2 + 260, from which
b2 = (16.4)2 − 260 = 8.96, or b = ±3 = +3 m.
The direction cosines are
cos θx = 316.4 = 0.1829, cos θy = −8
16.4 = −0.4878,
cos θz = 1416.4 = 0.8537
Problem 2.84 Observers at A and B use theodolitesto measure the direction from their positions to a rocketin flight. If the coordinates of the rocket’s position at agiven instant are (4, 4, 2) km, determine the directioncosines of the vectors rAR and rBR that the observerswould measure at that instant.
B (5,0,2) km
A
rAR
rBR
x
y
z
Solution: The vector rAR is given by
rAR = 4i + 4j + 2k km
and the magnitude of rAR is given by
|rAR| =√
(4)2 + (4)2 + (2)2 km = 6 km.
The unit vector along AR is given by
uAR = rAR/|rAR|.Thus, uAR = 0.667i + 0.667j + 0.333k
and the direction cosines are
cos θx = 0.667, cos θy = 0.667, and cos θz = 0.333.
The vector rBR is given by
rBR = (xR − xB)i + (yR − yB)j + (zR − zB)k km
= (4 − 5)i + (4 − 0)j + (2 − 2)k km
and the magnitude of rBR is given by
|rBR| =√
(1)2 + (4)2 + (0)2 km = 4.12 km.
The unit vector along BR is given by
eBR = rBR/|rBR|.Thus, uBR = −0.242i + 0.970j + 0k
and the direction cosines are
cos θx = −0.242, cos θy = 0.970, and cos θz = 0.0.
B (5,0,2) km
A
rAR rBRx
y
z
Full file at https://fratstock.euProblem 2.85 In Problem 2.84, suppose that the co-ordinates of the rocket’s position are unknown. At agiven instant, the person at A determines that the direc-tion cosines of rAR are cos θx = 0.535, cos θy = 0.802,and cos θz = 0.267, and the person at B determinesthat the direction cosines of rBR are cos θx = −0.576,cos θy = 0.798, and cos θz = −0.177. What are thecoordinates of the rocket’s position at that instant.
Solution: The vector from A to B is given by
rAB = (xB − xA)i + (yB − yA)j + (zB − zA)k or
rAB = (5 − 0)i + (0 − 0)j + (2 − 0)k = 5i + 2k km.
The magnitude of rAB is given by |rAB | =√
(5)2 + (2)2 =5.39 km. The unit vector along AB, uAB , is given by
uAB = rAB/|rAB | = 0.928i + 0j + 0.371k km.
The unit vector along the line AR,
uAR = cos θxi + cos θyj + cos θzk = 0.535i + 0.802j + 0.267k.
Similarly, the vector along BR, uBR = −0.576i+0.798−0.177k.From the diagram in the problem statement, we see thatrAR = rAB+rBR. Using the unit vectors, the vectors rAR and rBR can be writtenas
rAR = 0.535rARi + 0.802rARj + 0.267rARk, and
rBR = −0.576rBRi + 0.798rBRj − 0.177rBRk.
Substituting into the vector addition rAR = rAB +rBR and equatingcomponents, we get, in the x direction, 0.535rAR = −0.576rBR,and in the y direction, 0.802rAR = 0.798rBR. Solving, we get thatrAR = 4.489 km. Calculating the components, we get
rAR = rAReAR = 0.535(4.489)i + 0.802(4.489)j + 0.267(4.489)k.
Hence, the coordinates of the rocket, R, are (2.40, 3.60, 1.20) km.
Full file at https://fratstock.euProblem 2.86 The height of Mount Everest was origi-nally measured by a surveyor using the following proce-dure. He first measured the distance between two pointsA andB of equal altitude. Suppose that they are 10,000 ftabove sea level and are 32,000 ft apart. He then used atheodolite to measure the direction cosines of the vectorsfrom point A to the top of the mountain P and from pointB to P . Suppose that for rAP , the direction cosines arecos θx = 0.509, cos θy = 0.509, cos θz = 0.694, andfor rBP they are cos θx = −0.605, cos θy = 0.471,cos θz = 0.642. The z axis of the coordinate system isvertical. What is the height of Mount Everest above sealevel?
P
y
A
z
B x
Solution: Construct the two triangles: (a) Triangle ABP , whichhas one known side, AB, and two known adjacent interior angles θA
and θB (b) Triangle AOP , which is a right triangle with a derivedknown interior angle θAO . From triangle ABP , determine the lengthof AP , and from triangle APO and the derived interior angle, deter-mine the height above the base, OP . The interior angles of the triangleABP are
θA = θAX = cos−1(0.509) = 59.4◦,
θB = 180 − θBX = 180 − cos−1(−0.605)
= 180◦ − 127.2◦ = 52.77◦ and
β = 180 − 59.4 − 52.77 = 67.83◦.
From the law of sines:
|rAB |sin 67.83◦ =
|rBP |sin 59.4◦ =
|rAP |sin 52.77◦ .
Therefore the length of side AP is
|rAP | = |rAB |(
sin 52.77sin 67.83
)= 32000(0.8598) = 27512.9 ft.
The interior angle of the triangle APO is θAO = 90 − θAZ =90 − cos−1(0.694) = 90 − 46.05 = 43.95◦. Therefore the lengthof the side OP is hOP = |rAP | sin 43.95◦ = 27512.9(0.6940) =19093.9 ft. Check: The z-component of |rAP | is hop =|rAP | sin θAZ = 19093.9 ft. check.The base is 10000 ft above sea level, hence the height of P above sealevel is
P = 19093.9 + 10000 = 29094 ft
A
xy
z
B
P
A
O
xy
z
B
P
θA
θ B
θ AZ β
Full file at https://fratstock.euProblem 2.87 The distance from point O to point Ais 20 ft. The straight line AB is parallel to the y axis,and point B is in the x-z plane. Express the vector rOA
in terms of scalar components.Strategy: You can resolve rOA into a vector from Oto B and a vecotr from B to A. You can then resolve thevector form O to B into vector components parallel tothe x and z axes. See Example 2.9.
y
x
z
30°
B
O
A
rOA
60°
Solution: See Example 2.10. The length BA is, from the righttriangle OAB,
|rAB | = |rOA| sin 30◦ = 20(0.5) = 10 ft.
Similarly, the length OB is
|rOB | = |rOA| cos 30◦ = 20(0.866) = 17.32 ft
The vector rOB can be resolved into components along the axes bythe right triangles OBP and OBQ and the condition that it lies in thex-z plane.Hence,
rOB = |rOB |(i cos 30◦ + j cos 90◦ + k cos 60◦) or
rOB = 15i + 0j + 8.66k.
The vector rBA can be resolved into components from the conditionthat it is parallel to the y-axis. This vector is
rBA = |rBA|(i cos 90◦ + j cos 0◦ + k cos 90◦) = 0i + 10j + 0k.
The vector rOA is given by rOA = rOB + rBA, from which
rOA = 15i + 10j + 8.66k (ft)
P
rOA
B
QO
A
z
y
x
60°
30°
Problem 2.88 The magnitude of r is 100 in. The strai-ght line from the head of r to point A is parallel to the xaxis, and point A is contained in the y-z plane. Expressr in terms of scalar components.
BedfordFalls
BedfordFalls
45°60°
Ox
y
r
z
A
Solution: The vector r can be expressed as the sum of the twovectors, r = rOA+rAP , both of which can be resolved into directioncosine components. The magnitudes can be determined from the lawof sines for the triangle OAP .
rOA = |rOA|(i cos 90◦ + j cos 30◦ + k cos 60◦)
rOA = |rOA|(0i + 0.866j + 0.5k). Similarly,
rAP = |rAP |(i cos 0 + j cos 90 + k cos 90) = |rAP |(1i + 0j + 0k)
Since rAP is parallel to the x-axis, it makes an angle of 90◦ with they-z plane, and the triangle OAP is a right triangle. From the law ofsines
|r|sin 90◦ =
|rAP |sin 45◦ =
|rOA|sin 45◦ ,
from which |rAP | = |rOA| = 100(0.707) = 70.7. Substitutingthese values into the vectors
r = rOA + rAP = 70.7(1i + 0.866j + 0.5k)
= 70.7i + 61.2j + 35.4k (in.)
y
x
z
A P
45°60°
r
Full file at https://fratstock.euProblem 2.89 The straight line from the head of F topoint A is parallel to the y axis, and point A is containedin the x-z plane. The x component of F is Fx = 100 N.(a) What is the magnitude of F?.(b) Determine the angles θx, θy , and θz between F and
the positive coordinate axes.
y
z
x
A
F
20°
60°
O
Solution: The triangle OpA is a right triangle, since OA lies inthe x-z plane, and Ap is parallel to the y-axis. Thus the magnitudesare given by the sine law:
|rAp|sin 20◦ =
|F|sin 90◦ =
|rOA|sin 70◦ ,
thus |rAp| = |F|(0.342) and |rOA| = |F|(0.9397). The compo-nents of the two vectors are from the geometry
rOA = |rOA|(i cos 30◦ + j cos 90◦ + k cos 60◦)
= |rOA|(0.866i + 0j + 0.5k) and
rAp = |rAp|(i cos 90◦ + j cos 0◦ + k cos 90◦) = |rAp|(0i + 1j + 0k)
Noting F = rOA + rAp, then from above
F = |F|(0.3420)(0i + 1j + 0k) + |F|(0.9397)(0.866i + 0j + 0.5k)
F = |F|(0.8138i + 0.342j + 0.4699k)
The x-component is given to be 100 N. Thus,
(a) |F| =100
0.8138= 122.9 N The angles are given by
(b) θx = cos−1(0.8138) = 35.5◦ ,
θy = cos−1(0.342) = 70◦ and θz = cos−1(0.4699) = 62◦
y
z
x
A
O
60°
20°F
q
P
Problem 2.90 The position of a point P on the sur-face of the earth is specified by the longitude λ, measuredfrom the point G on the equator directly south of Green-wich, England, and the latitude L measured from theequator. Longitude is given as west (W) longitude or east(E) longitude, indicating whether the angle is measuredwest or east from point G. Latitude is given as north (N)latitude or south (S) latitude, indicating whether the an-gle is measured north or south from the equator. Supposethat P is at longitude 30◦ W and latitude 45◦ N. Let RE
be the radius of the earth. Using the coordinate systemshown, determine the components of the position vectorof P relative to the center of the earth. (Your answer willbe in terms of RE .)
G
x
y
N
Equator
z
λO
L
P
Solution: Drop a vertical line from pointP to the equatorial plane.Let the intercept be B (see figure). The vector position of P is thesum of the two vectors: P = rOB + rBP . The vector rOB =|rOB |(i cos λ + 0j + k sin λ). From geometry, the magnitude is|rOB | = RE cos θ.The vector rBP = |rBP |(0i + 1j + 0k). From geometry, themagnitude is |rBP | = RE sin θP . Substitute: P = rOB +rBP = RE(i cos λ cos θ + j sin θ + k sin λ cos θ). Substitutefrom the problem statement: λ = +30◦, θ = 45◦. Hence
P = RE(0.6124i + 0.707j + 0.3536k)
y
z
x
P
B
λ
O
Full file at https://fratstock.euProblem 2.91 An engineer calculates that the magni-tude of the axial force in one of the beams of a geodesicdome is |P = 7.65 kN. The cartesian coordinates ofthe endpoints A and B of the straight beam are (−12.4,22.0, −18.4) m and (−9.2, 24.4, −15.6) m, respectively.Express the force P in terms of scalar components.
B
P
A
Solution: The components of the position vector from B to A are
rBA = (Ax − Bx)i + (Ay − By)j + (Az − Bz)k
= (−12.4 + 9.2)i + (22.0 − 24.4)j
+(−18.4 + 15.6)k
= −3.2i − 2.4j − 2.8k (m).
Dividing this vector by its magnitude, we obtain a unit vector that points fromB toward A:
eBA = −0.655i − 0.492j − 0.573k.
Therefore
P = |P|eBA
= 7.65 eBA
= −5.01i − 3.76j − 4.39k (kN).
Problem 2.92 The cable BC exerts an 8-kN force Fon the bar AB at B.(a) Determine the components of a unit vector that
points from B toward point C.(b) Express F in terms of components.
yB (5, 6, 1) m
A
C (3, 0, 4) m
z
x
F
Solution:
(a) eBC =rBC
|rBC | =(xC − xB)i + (yC − yB)j + (zC − zB)k√(xC − xB)2 + (yC − yB)2 + (zC − zB)2
eBC =−2i − 6j + 3k√
22 + 62 + 32= −2
7i − 6
7j +
37k
eBC = −0.286i − 0.857j + 0.429k
(b) F = |F|eBC = 8eBC = −2.29i − 6.86j + 3.43k (kN)
x
yB (5, 6, 1) m
C (3, 0, 4) m
z
A
F
Full file at https://fratstock.euProblem 2.93 A cable extends from point C to pointE. It exerts a 50-lb force T on plate C that is directedalong the line from C to E. Express T in terms of scalarcomponents.
Dx
CB
A
20°4 ft
4 ft
6 ft
2 ft
E
z
y
T
Solution: Find the unit vector eCE and multiply it times the mag-nitude of the force to get the vector in component form,
eCE =rCE
|rCE | =(xE − xC)i + (yE − yC)j + (zE − zC)k√(xE − xC)2 + (yE − yC)2 + (zE − zC)2
The coordinates of point C are (4, −4 sin 20◦, 4 cos 20◦) or(4, −1, 37, 3.76) (ft) The coordinates of point E are (0, 2, 6) (ft)
eCE =(0 − 4)i + (2 − (−1.37))j + (6 − 3.76)k√
42 + 3.372 + 2.242
eCE = −0.703i + 0.592j + 0.394k
T = 50eCE (lb)
T = −35.2i + 29.6j + 19.7k (lb)
D x
CB
A
20°4 ft
4 ft
6 ft
2 ft
E
z
y
T
T
Problem 2.94 What are the direction cosines of theforce T in Problem 2.93?
Solution: From the solution to Problem 2.93,
eCE = −0.703i + 0.592j + 0.394k
However
eCE = cos θxi + cos θyj + cos θzk
Hence,
cos θx = −0.703
cos θy = 0.592
cos θz = 0.394
Full file at https://fratstock.euProblem 2.95 The cable AB exerts a 200-lb forceFAB at point A that is directed along the line from A toB. Express FAB in terms of scalar components.
A (6, 0, 10) ft
B
C8 ft
x
z
y
6 ft
FABFAC
8 ft
Solution: The coordinates of B are B(0,6,8). The position vectorfrom A to B is
rAB = (0 − 6)i + (6 − 0)j + (8 − 10)k = −6i + 6j − 2k
The magnitude is |rAB | =√
62 + 62 + 22 = 8.718 ft.The unit vector is
uAB =−6
8.718i +
68.718
j − 28.718
k
FAB = |FAB |uAB = 200(−0.6882i + 0.6882j − 0.2294k) or
uAB = −0.6882i + 0.6882j − 0.2294k.
The components of the force are
FAB = |FAB |uAB = 200(−0.6882i + 0.6882j − 0.2294k) or
FAB = −137.6i + 137.6j − 45.9k
A(6, 0, 10) ft
B
C8 ft
x
z
y
6 ft
8 ft
Problem 2.96 Consider the cables and wall describedin Problem 2.95. Cable AB exerts a 200-lb force FAB
at point A that is directed along the line from A to B.The cable AC exerts a 100-lb force FAC at point A thatis directed along the line from A to C. Determine themagnitude of the total force exerted at point A by thetwo cables.
Solution: Refer to the figure in Problem 2.81. From Problem 2.81the force FAB is
FAB = −137.6i + 137.6j − 45.9k
The coordinates of C are C(8,6,0). The position vector from A to C is
rAC = (8 − 6)i + (6 − 0)j + (0 − 10)k = 2i + 6j − 10k.
The magnitude is |rAC | =√
22 + 62 + 102 = 11.83 ft.The unit vector is
uAC =2
11.83i +
611.83
j − 1011.83
k = 0.1691i + 0.5072j − 0.8453k.
The force is
FAC = |FAC |uAC = 100uAC = 16.9i + 50.7j − 84.5k.
The resultant of the two forces is
FR = FAB + FAC = (−137.6 + 16.9)i + (137.6 + 50.7)j
+ (−84.5 − 45.9)k.
FR = −120.7i + 188.3j − 130.4k.
The magnitude is
|FR| =√
120.72 + 188.32 + 130.42 = 258.9 lb
Full file at https://fratstock.euProblem 2.97 The 70-m-tall tower is supported bythree cables that exert forces FAB , FAC , and FAD on it.The magnitude of each force is 2 kN. Express the totalforce exerted on the tower by the three cables in termsof scalar components.
A
x
y
40 m
60 m
40 m40 m
60 m
B
C
D
z
FABFAC
FAD
A
Solution: The coordinates of the points are A (0, 70, 0), B (40,0, 0), C (−40, 0, 40) D (−60, 0, −60).The position vectors corresponding to the cables are:
rAD = (−60 − 0)i + (0 − 70)j + (−60 − 0)k
rAD = −60i − 70k − 60k
rAC = (−40 − 0)i + (0 − 70)j + (40 − 0)k
rAC = −40i − 70j + 40k
rAB = (40 − 0)i + (0 − 70)j + (0 − 0)k
rAB = 40i − 70j + 0k
The unit vectors corresponding to these position vectors are:
uAD =rAD
|rAD| =−60110
i − 70110
j − 60110
k
= −0.5455i − 0.6364j − 0.5455k
uAC =rAC
|rAC | = −4090
i − 7090
j +4090
k
= −0.4444i − 0.7778j + 0.4444k
uAB =rAB
|rAB | =40
80.6i − 70
80.6j + 0k = 0.4963i − 0.8685j + 0k
The forces are:
FAB = |FAB |uAB = 0.9926i − 1.737j + 0k
FAC = |FAC |uAC = −0.8888i − 1.5556j + 0.8888
FAD = |FAD|uAD = −1.0910i − 1.2728j − 1.0910k
The resultant force exerted on the tower by the cables is:
FR = FAB + FAC + FAD = −0.9872i − 4.5654j − 0.2022k kN
A
x
A
40 m
60 m
40 m 40 m
60 m
B
C
D
E
FABFAC
FAD
A
Full file at https://fratstock.euProblem 2.98 Consider the tower described in Pro-blem 2.97. The magnitude of the force FAB is 2 kN.The x and z components of the vector sum of the forcesexerted on the tower by the three cables are zero. Whatare the magnitudes of FAC and FAD?
Solution: From the solution of Problem 2.83, the unit vectors are:
uAC =rAC
|rAC | = −4090
i − 7090
j +4090
k
= −0.4444i − 0.7778j + 0.4444k
uAD =rAD
|rAD| =−60110
i − 70110
j − 60110
= −0.5455i − 0.6364j − 0.5455k
From the solution of Problem 2.83 the force FAB is
FAB = |FAB |uAB = 0.9926i − 1.737j + 0k
The forces FAC and FAD are:
FAC = |FAC |uAC = |FAC |(−0.4444i − 0.7778j + 0.4444k)
FAD = |FAD|uAD = |FAD|(−0.5455i − 0.6364j − 0.5455k)
Taking the sum of the forces:
FR = FAB + FAC + FAD = (0.9926 − 0.4444|FAC | − 0.5455|FAD|)i+(−1.737 − 0.7778|FAC | − 0.6364|FAD|)j+(0.4444|FAC | − 0.5455|FAD|)k
The sum of the x- and z-components vanishes, hence the set of simultaneousequations:
0.4444|FAC | + 0.5455|FAD| = 0.9926 and
0.4444|FAC | − 0.5455|FAD| = 0
These can be solved by means of standard algorithms, or by the use of com-mercial packages such as TK Solver Plus ® or Mathcad®. Here a hand heldcalculator was used to obtain the solution:
|FAC | = 1.1168 kN |FAD| = 0.9098 kN
Problem 2.99 Express the position vector from pointO to the collar at A in terms of scalar components.
Ox
y
A
6 ft
7 ft
4 ft
4 ftz
Solution: The vector from O to A can be expressed as the sumof the vectors rOT from O to the top of the slider bar, and rTA fromthe top of the slider bar to A. The coordinates of the top and base ofthe slider bar are: T (0, 7, 0), B (4, 0, 4). The position vector of thetop of the bar is: rOT = 0i + 7j + 0k. The position vector from thetop of the bar to the base is:
rTB = (4 − 0)i + (0 − 7)j + (4 − 0)k. or
rTB = 4i − 7j + 4k. The unit vector pointing from the top of thebar to the base is
uTB =rTB
|rTB | =49i − 7
9j +
49k = 0.4444i − 0.7778j + 0.4444k.
The collar position is
rTA = |rTA|uTB = 6(0.4444i − 0.7778j + 0.4444k)
= 2.6667i − 4.6667j + 2.6667,
measured along the bar. The sum of the two vectors is the positionvector of A from origin O:
rOA = (2.6667 + 0)i + (−4.6667 + 7)j + (2.6667 + 0)k
= 2.67i + 2.33j + 2.67k ft
O
A
6 ft
7 ft
4 ft
4 ft
Full file at https://fratstock.euProblem 2.100 The cable AB exerts a 32-lb force Ton the collar at A. Express T in terms of scalar compo-nents.
x
y
A
6 ft
B
4 ft
4 ft
7 ft
4 ft
z
T
Solution: The coordinates of point B are B (0, 7, 4). The vectorposition of B is rOB = 0i + 7j + 4k.The vector from point A to point B is given by
rAB = rOB − rOA.
From Problem 2.86, rOA = 2.67i + 2.33j + 2.67k. Thus
rAB = (0 − 2.67)i + (7 − 2.33)j + (4 − 2.67)j
rAB = −2.67i + 4.67j + 1.33k.
The magnitude is
|rAB | =√
2.672 + 4.672 + 1.332 = 5.54 ft.
The unit vector pointing from A to B is
uAB =rAB
|rAB | = −0.4819i + 0.8429j + 0.2401k
The force T is given by
TAB = |TAB |uAB = 32uAB = −15.4i + 27.0j + 7.7k (lb)
x
y
A
6 ftB
4 ft
4 ft
7 ft
4 ft
z
Full file at https://fratstock.euProblem 2.101 The circular bar has a 4-m radius andlies in the x-y plane. Express the position vector frompoint B to the collar at A in terms of scalar components.
z
y
x
A
B
4 m
4 m
3 m
20°
Solution: From the figure, the point B is at (0, 4, 3) m. The coordinates ofpoint A are determined by the radius of the circular bar and the angle shown in thefigure. The vector from the origin to A is rOA = 4 cos(20◦)i+4 sin(20◦)j m.Thus, the coordinates of point A are (3.76, 1.37, 0) m. The vector from Bto A is given by rBA = (xA − xB)i + (yA − yB)j + (zA − zB)k =3.76i − 2.63j − 3k m. Finally, the scalar components of the vector from B toA are (3.76, −2.63, −3) m.
z
y
x
A
B
4ft
4 ft
3 ft
20°
Problem 2.102 The cable AB in Problem 2.101 exertsa 60-N force T on the collar at A that is directed alongthe line from A toward B. Express T in terms of scalarcomponents.
Solution: We know rBA = 3.76i − 2.63j − 3k m from Problem 2.101.The unit vector uAB = −rBA/|rBA|. The unit vector is uAB = −0.686i+0.480j + 0.547k. Hence, the force vector T is given by
T= |T|(−0.686i+ 0.480j+ 0.547k) N=−41.1i + 28.8j + 32.8k N
Problem 2.103 Determine the dot product of the vec-tors
U = 8i − 6j + 4k
and V = 3i + 7j + 9k.
Solution:
U · V = UxVx + UyVy + UzVz
= (8)(3) + (−6)(7) + (4)(9)
U · V = 18
Problem 2.104 Determine the dot product U · V ofthe vectors U = 40i+20j+60k and V = −30i+15k.
Solution: Use Eq. 2.23.
U · V = (40)(−30) + (20)(0) + (15)(60) = −300
Problem 2.105 What is the dot product of the positionvector r = −10i + 25j (m) and the force
F = 300i + 250j + 300k (N)?
Solution: Use Eq. (2.23).
F · r = (300)(−10) + (250)(25) + (300)(0) = 3250 N-m
Problem 2.106 What is the dot product of the positionvector r = 4i − 12j − 3k (ft) and the force F = 20i +30j − 10k (lb)?
Solution: Use Eq. (2.23).
r · F = 4(20) + 30(−12) − 10(−3) = −250 ft lb
Full file at https://fratstock.euProblem 2.107 Two perpendicular vectors are givenin terms of their components by
U = Uxi − 4j + 6k
and V = 3i + 2j − 3k.
Use the dot product to determine the component Ux.
Solution: When the vectors are perpendicular, U · V ≡ 0.Thus
U · V = UxVx + UyVy + UzVz = 0
= 3Ux + (−4)(2) + (6)(−3) = 0
3Ux = 26
Ux = 8.67
Problem 2.108 Three vectors
U = Uxi + 3j + 2k
V = −3i + Vyj + 3k
W = −2i + 4j + Wzk
are mutually perpendicular. Use the dot product to de-termine the components Ux, Vy , and Wz
Solution: For mutually perpendicular vectors, we have three equations,i.e.,
U · V = 0
U · W = 0
V · W = 0
Thus
−3Ux + 3Vy + 6 = 0−2Ux + 12 + 2Wz = 0+6 + 4Vy + 3Wz = 0
3 Eqns3 Unknowns
Solving, we get
Ux = 2.857Vy = 0.857Wz = −3.143
Problem 2.109 The magnitudes |U| = 10 and |V| =20.(a) Use the definition of the dot product to determine
U · V.(b) Use Eq. (2.23) to obtain U · V.
x
y
V
45°
U
30°
Solution:(a) The definition of the dot product (Eq. (2.18)) is
U · V = |U||V| cos θ. Thus
U · V = (10)(20) cos(45◦ − 30◦) = 193.2
(b) The components of U and V are
U = 10(i cos 45◦ + j sin 45◦) = 7.07i + 7.07j
V = 20(i cos 30◦ + j sin 30◦) = 17.32i + 10j
From Eq. (2.23)
U · V=(7.07)(17.32) + (7.07)(10) = 193.2
U
V
y
x30°
45°
Full file at https://fratstock.euProblem 2.110 By evaluating the dot product U · V,prove the identity cos(θ1 − θ2) = cos θ1 cos θ2 +sin θ1 sin θ2.Strategy: Evaluate the dot product both by using thedefinition and by using Eq. (2.23).
x
y
V
θ1
θ2
U
Solution: The strategy is to use the definition Eq. (2.18) and theEq. (2.23). From Eq. (2.18) and the figure,
U · V = |U||V| cos(θ1 − θ2). From Eq. (2.23) and the figure,
U = |U|(i cos θ1 + j sin θ2), V = |V|(i cos θ2 + j sin θ2),
and the dot product isU·V = |U||V|(cos θ1 cos θ2+sin θ1 sin θ2).Equating the two results:
U · V = |U||V| cos(θ1 − θ2) = |U||V|(cos θ1 cos θ2 + sin θ1 sin θ2),
from which if |U| �= 0 and |V| �= 0, it follows that
cos(θ1 − θ2) = cos θ1 cos θ2 + sin θ1 sin θ2 , Q.E.D.
y
θ1
θ2
x
U
V
Problem 2.111 Use the dot product to determine theangle between the forestay (cable AB) and the backstay(cable BC) of the sailboat in Problem 2.41.
Solution: The unit vector from B to A is
eBA =rBA
|rBA| = −0.321i − 0.947j
The unit vector from B to C is
eBC =rBC
|rBC | = 0.385i − 0.923j
From the definition of the dot product, eBA · eBC = 1 · 1 · cos θ,where θ is the angle between BA and BC. Thus
cos θ = (−0.321)(0.385) + (−0.947)(−0.923)
cos θ = 0.750496
θ = 41.3◦
y
x
B (4,13) m
C(9,1) m
A(0,1.2) m
Full file at https://fratstock.euProblem 2.112 What is the angle θ between the str-aight lines AB and AC?
A(8, 6, 4) ft
(–4, 5, –4) ft
(6, 0, 6) ft
B
C
θ
x
y
z
Solution: From the given coordinates, the position vectors are:
rOB = −4i + 5j − 4k, rOA = 8i + 6j + 4k, and
rOC = 6i + 0j + 6k.
The straight lines correspond to the vectors:
rAB = rOB − rOA = −12i − j − 8k,
rAC = rOC − rAC = −2i − 6j + 2k
The dot product is given by
rAB · rAC = (−2)(−12) + (−1)(−6) + (+2)(−8) = 14.
The magnitudes of the vectors are:
|rAC =√
22 + 62 + 22| = 6.6333, and
|rAB =√
122 + 12 + 82| = 14.456.
From the definition of the dot product, the angle is
cos θ =rAC · rAB
|rAC ||rAB | =14
(14.456)(6.633)= 0.1460.
Take the principal value: θ = 81.6◦
A (8, 6, 4)
(–4, 5, –4)
(6, 0, 6)
B
C
θ
x
y
z
Problem 2.113 The ship O measures the positions ofthe ship A and the airplane B and obtains the coordinatesshown. What is the angle θ between the lines of sightOA and OB?
x
y
z
A
B
θ
O
(6, 0, 3) km
(4, 4, –4) km
Solution: From the coordinates, the position vectors are:
rOA = 6i + 0j + 3k and rOB = 4i + 4j − 4k
The dot product: rOA · rOB = (6)(4) + (0)(4) + (3)(−4) = 12
The magnitudes: |rOA| =√
62 + 02 + 32 = 6.71 km and
|rOA| =√
42 + 42 + 42 = 6.93 km.
From Eq. (2.24) cos θ = rOA·rOB|rOA||rOB | = 0.2581, from which θ =
±75◦. From the problem and the construction, only the positive angle
makes sense, hence θ = 75◦
x
y
zA
B
θO
(6, 0, 3) km
(4, 4, –4) km
Full file at https://fratstock.euProblem 2.114 Astronauts on the space shuttle useradar to determine the magnitudes and direction cosinesof the position vectors of two satellites A and B. Thevector rA from the shuttle to satellite A has magnitude2 km and direction cosines cos θx = 0.768, cos θy =0.384, cos θz = 0.512. The vector rB from the shuttleto satellite B has magnitude 4 km and direction cosinescos θx = 0.743, cos θy = 0.557, cos θz = −0.371.What is the angle θ between the vectors rA and rB?
x
rB
z
B
A rA
y
θ
Solution: The direction cosines of the vectors along rA and rB
are the components of the unit vectors in these directions (i.e., uA =cos θxi + cos θyj + cos θzk, where the direction cosines are thosefor rA). Thus, through the definition of the dot product, we can findan expression for the cosine of the angle between rA and rB .
cos θ = cos θxA cos θxB + cos θyA cos θyB + cos θzA cos θzB .
Evaluation of the relation yields
cos θ = 0.594 ⇒ θ = 53.5◦ .
x
rB
z
B
A rAy
θ
Problem 2.115 The cable BC exerts an 800-N forceF on the bar AB at B. Use Eq. (2.26) to determine thevector component of F parallel to the bar.
yB (5, 6, 1) m
A
C (3, 0, 4) m
z
x
F
Solution: Eqn. 2.26 is UP = (e · U)e where U is the vectorfor which you want the component parallel to the direction indicatedby the unit vector e.For the problem at hand, we must find two unit vectors. We needeBC to be able to write the force F(F = |F|eBC) and eBA ∼ thedirection parallel to the bar.
eBC =rBC
|rBC | =(xC − xB)i + (yC − yB)j + (zC − zB)k√(xC − xB)2 + (yC − yB)2 + (zC − zB)2
eBC =(3 − 5)i + (0 − 6)j + (4 − 1)k√
22 + 62 + 32
eBC = −27i − 6
7j +
37k
Similarly
eBA =−5i − 6j − 1k√
52 + 62 + 12
eBA = −0.635i − 0.762j − 0.127k
Now F = |F|eBC = 800 eBC
F = −228.6i − 685.7j + 342.9k N
FP = (F · eBA)eBA
FP = (624.1)eBA
FP = −396.3i − 475.6j − 79.3k N
B (5, 6, 1) m
A
C (3, 0, 4) m
z
x
F
Full file at https://fratstock.euProblem 2.116 The force F = 21i+14j (kN). Resol-ve it into vector components parallel and normal to theline OA.
A (6, –2, 3) m
x
y
z
F
O
Solution: The position vector of point A is
rA = 6i − 2j + 3k
The magnitude is |rA| =√
62 + 22 + 32 = 7. The unit vectorparallel to OA is eOA = rA
|rA| = 67 i − 2
7 j + 37k
(a) The component of F parallel to OA is
(F · eOA) eOA = ((3)(6) + (−2)(2))(
17
)(6i − 2j + 3k)
FP = 12i − 4j + 6k (kN)
(b) The component of F normal to OA is
FN = F − Fp = (21 − 12)i + (14 − (−4))j + (0 − 6)k
= 9i + 18j − 6k (kN)
A(6, –2, 3) m
x
y
z
F
Problem 2.117 At the instant shown, the Harrier’sthrust vector is T = 3800i + 15,300j − 1800k (lb), andits velocity vector is v = 24i+6j−2k (ft/s). Resolve Tinto vector components parallel and normal to v. (Theseare the components of the airplane’s thrust parallel andnormal to the direction of its motion.)
x
y
T
v
Solution: The magnitude of the velocity vector is given by
|v| =√
v2x + v2
y + v2z =
√242 + 62 + (−2)2.
Thus, |v| = 24.8 ft/s. The components of the unit vector in thedirection of the velocity vector are given by
ex =vx
|v| , ey =vy
|v| , and ez =vz
|v| .
Substituting numerical values, we get ex = 0.967, ey = 0.242, andez = −0.0806. The dot product of T and this unit vector gives thecomponent of T parallel to the velocity. The resulting equation isTparallel = Txex + Tyey + Tzez . Substituting numerical values,
we get Tparallel = 7232.12 lb. The magnitude of the vector T is
15870 lb. Using the Pythagorean Theorem, we get
Tnormal =√|T |2 − (Tparallel)2 = 14130 lb.
x
y
T
v
Full file at https://fratstock.euProblem 2.118 Cables extend from A to B and fromA to C. The cable AC exerts a 1000-lb force F at A.(a) What is the angle between the cables AB and AC?(b) Determine the vector component of F parallel to the
cable AB.
F
A (0, 7, 0) ft
B
C
x
y
z (14, 0, 14) ft(0, 0, 10) ft
Solution: Use Eq. (2.24) to solve.(a) From the coordinates of the points, the position vectors are:
rAB = (0 − 0)i + (0 − 7)j + (10 − 0)k
rAB = 0i − 7j + 10k
rAC = (14 − 0)i + (0 − 7)j + (14 − 0)k
rAC = 14i − 7j + 14k
The magnitudes are:
|rAB | =√
72 + 102 = 12.2 (ft) and
|rAB | =√
142 + 72 + 142 = 21.
The dot product is given by
rAB · rAC = (14)(0) + (−7)(−7) + (10)(14) = 189.
The angle is given by
cos θ =189
(12.2)(21)= 0.7377,
from which θ = ±42.5◦. From the construction: θ = +42.5◦(b) The unit vector associated with AB is
eAB =rAB
|rAB | = 0i − 0.5738j + 0.8197k.
The unit vector associated with AC is
eAC =rAC
|rAC | = 0.6667i − 0.3333j + 0.6667k.
Thus the force vector along AC is
FAC = |F|eAC = 666.7i − 333.3j + 666.7k.
The component of this force parallel to AB is
(FAC · eAB)eAB = (737.5)eAB = 0i − 423.2j + 604.5k (lb)
A (0, 7, 0) ft
B
C
x
y
z(14, 0, 14) ft
(0, 0, 10) ft
Problem 2.119 Consider the cables AB and ACshown in Problem 2.118. Let rAB be the position vectorfrom point A to point B. Determine the vector compo-nent of rAB parallel to the cable AC.
Solution: From Problem 2.100, rAB = 0i − 7j + 10k, and eAC =0.6667i−0.3333j+0.6667k. Thus rAB ·eAC = 9, and (rAB ·eAC)eAC =6i − 3j + 6k
Full file at https://fratstock.euProblem 2.120 The force F = 10i + 12j − 6k (N).Determine the vector components of F parallel and nor-mal to line OA.
y
x
z
(0, 6, 4) m
O
A
F
Solution: Find eOA = rOA|rOA|
Then
FP = (F · eOA)eOA
and FN = F − FP
eOA =0i + 6j + 4k√
62 + 42=
6j + 4k√52
eOA =6
7.21j +
47.21
k = 0.832j + 0.555k
FP = [(10i + 12j − 6k) · (0.832j + 0.555k)]eOA
FP = [6.656]eOA = 0i + 5.54j + 3.69k (N)
FN = F − FP
FN = 10i + (12 − 5.54)j + (−6 − 3.69k)
FN = 10i + 6.46j − 9.69k N
y
x
z
(0, 6, 4) m
O
A
F
Problem 2.121 The rope AB exerts a 50-N force T oncollar A. Determine the vector component of T parallelto bar CD.
0.4 m
0.5 m
0.15 m
0.3 m0.2 m
0.25 m
0.2 mz
x
y
A
B C
D
O
T
Solution: The vector from C to D is rCD = (xD − xC)i +(yD − yC)j + (zD − zC)k. The magnitude of the vector
|rCD| =√
(xD − xC)2 + (yD − yC)2 + (zD − zC)2.
The components of the unit vector along CD are given by uCDx =(xD − xC)/|rCD|, uCDy = (yD − yC)/|rCD|, etc. Numericalvalues are |rCD| = 0.439 m, uCDx = −0.456, uCDy = −0.684,and uCDz = 0.570. The coordinates of point A are given by xA =xC + |rCA|eCDx, yA = yC + |rCA|uCDy , etc. The coordinatesof point A are (0.309, 0.163, 0.114) m. The vector from A to B andthe corresponding unit vector are found in the same manner as fromC to D above. The results are |rAB | = 0.458 m, uABx = −0.674,uABy = 0.735, and uABz = 0.079. The force T is given byT = |T|uAB . The result is T = −33.7i + 36.7j + 3.93k N.The component of T parallel to CD is given
Tparallel = T • uCD = −7.52 N.
The negative sign means that the component of T parallel to CD pointsfrom D toward C (opposite to the direction of the unit vector from Cto D).
0.4 m
0.5 m
0.15 m
0.3 m0.2 m
0.25 m
0.2 mz
x
y
A
B
C
D
O
T
Full file at https://fratstock.euProblem 2.122 In Problem 2.121, determine the vec-tor component of T normal to the bar CD.
Solution: From the solution of Problem 2.121, |T | = 50 N, and the com-ponent of T parallel to bar CD is Tparallel = −7.52 N. The component of Tnormal to bar CD is given by
Tnormal =√|T|2 − (Tparallel)2 = 49.4 N.
Problem 2.123 The disk A is at the midpoint of thesloped surface. The string from A to B exerts a 0.2-lb force F on the disk. If you resolve F into vectorcomponents parallel and normal to the sloped surface,what is the component normal to the surface?
y
z
A
B
F
(0, 6, 0) ft
2 ft
8 ft
10 ft
x
Solution: Consider a line on the sloped surface from A perpendic-ular to the surface. (see the diagram above) By SIMILAR triangles wesee that one such vector is rN = 8j + 2k. Let us find the componentof F parallel to this line.The unit vector in the direction normal to the surface is
eN =rN
|rN | =8j + 2k√82 + 22
= 0.970j + 0.243k
The unit vector eAB can be found by
eAB =(xB − xA)i + (yB − yA)j + (zB − zA)h√(xB − xA)2 + (yB − yA)2 + (zB − zA)2
Point B is at (0, 6, 0) (ft) and A is at (5, 1, 4) (ft).Substituting, we get
eAB = −0.615i + 0.615j − 0.492k
Now F = |F|eAB = (0.2)eAB
F = −0.123i + 0.123j − 0.0984k (lb)
The component of F normal to the surface is the component parallelto the unit vector eN .
FNORMAL = (F · eN )eN = (0.955)eN
FNORMAL = 0i + 0.0927j + 0.0232k lb
y
z8
8
2
2
y
z
A
B
FC
(0, 6, 0) ft
2 ft
8 ft
10 ft
x
Problem 2.124 In Problem 2.123, what is the vectorcomponent of F parallel to the surface?
Solution: From the solution to Problem 2.123,
F = −0.123i + 0.123j − 0.0984k (lb) and
FNORMAL = 0i + 0.0927j + 0.0232k (lb)
The component parallel to the surface and the component normal tothe surface add to give F(F = FNORMAL + Fparallel).
Thus
Fparallel = F − FNORMAL
Substituting, we get
Fparallel = −0.1231i + 0.0304j − 0.1216k lb
Full file at https://fratstock.euProblem 2.125 An astronaut in a maneuvering unitapproaches a space station. At the present instant, thestation informs him that his position relative to the originof the station’s coordinate system is rG = 50i + 80j +180k (m) and his velocity is v = −2.2j − 3.6k (m/s).The position of the airlock is rA = −12i + 20k (m).Determine the angle between his velocity vector and theline from his position to the airlock’s position.
Solution: Points G and A are located at G: (50, 80, 180) m andA: (−12, 0, 20) m. The vector rGA is rGA = (xA −xG)i+(yA −yG)j+(zA − zG)k = (−12−50)i+(0−80)j+(20−180)k m.The dot product between v and rGA is v • rGA = |v||rGA| cos θ =vxxGA +vyyGA +vzzGA, where θ is the angle between v and rGA.
Substituting in the numerical values, we get θ = 19.7◦.
y
zx
A
G
Problem 2.126 In Problem 2.125, determine the vec-tor component of the astronaut’s velocity parallel to theline from his position to the airlock’s position.
Solution: The dot product v • rGA = vxxGA + vyyGA +vzzGA = 752 (m/s)2 and the component of v parallel to GA isvparallel = |v| cos θ where θ is defined as in Problem 2.125 above.
vparallel = (4.22)(0.941) = 3.96 m/s
Full file at https://fratstock.euProblem 2.127 Point P is at longitude 30◦W and lat-itude 45◦N on the Atlantic Ocean between Nova Scotiaand France. (See Problem 2.90.) Point Q is at longitude60◦E and latitude 20◦N in the Arabian Sea. Use the dotproduct to determine the shortest distance along the sur-face of the earth from P to Q in terms of the radius ofthe earth RE .Strategy: Use the dot product to detrmine the anglebetween the lines OP and OQ; then use the definitionof an angle in radians to determine the distance along thesurface of the earth from P to Q.
Equator
y
z
x
P
N
O
45°
30° 60°
G
20°
Q
Solution: The distance is the product of the angle and the radius ofthe sphere, d = REθ, where θ is in radian measure. From Eqs. (2.18)and (2.24), the angular separation of P and Q is given by
cos θ =(
P · Q|P||Q|
).
The strategy is to determine the angle θ in terms of the latitude andlongitude of the two points. Drop a vertical line from each point Pand Q to b and c on the equatorial plane. The vector position of P isthe sum of the two vectors: P = rOB + rBP . The vector rOB =|rOB |(i cos λP + 0j + k sin λP ). From geometry, the magnitudeis |rOB | = RE cos θP . The vector rBP = |rBP |(0i + 1j + 0k).From geometry, the magnitude is |rBP | = RE sin θP . Substituteand reduce to obtain:
P = rOB + rBP = RE(i cos λP cos θP + j sin θP + k sin λP cos θP ).
A similar argument for the point Q yields
Q = rOC + rCQ = RE(i cos λQ cos θQ + j sin θQ + k sin λQ cos θQ)
Using the identity cos2 β + sin2 β = 1, the magnitudes are
|P| = |Q| = RE
The dot product is
P · Q = R2E(cos(λP − λQ) cos θP cos θQ + sin θP sin θQ)
Substitute:
cos θ =P · Q|P||Q| = cos(λP − λQ) cos θP cos θQ + sin θP sin θQ
Substitute λP = +30◦, λQ = −60◦, θp = +45◦,θQ = +20◦, to obtain cos θ = 0.2418, or θ = 1.326 radians.Thus the distance is d = 1.326RE
y
x
30°
45°
60°20°
RE
θQ
P
G
b c
N
Problem 2.128 Determine the cross product U × Vof the vectors U = 8i− 6j+ 4k and V = 3i+ 7j+ 9k.Strategy: Sine the vectors are expressed in terms oftheir components, you can use Eq. (2.34) to determinetheir cross product.
Solution:
U × V =
∣∣∣∣∣i j k8 −6 43 7 9
∣∣∣∣∣= (−54 − 28)i + (12 − 72)j + (56 + 18)k
U × V = −82i − 60j + 74k
Full file at https://fratstock.euProblem 2.129 Two vectors U = 3i + 2j and V =2i + 4j.(a) What is the cross product U × V?(b) What is the cross product V × U?
Solution: Use Eq. (2.34) and expand into 2 by 2 determinants.
U × V =
∣∣∣∣∣i j k3 2 02 4 0
∣∣∣∣∣ = i((2)(0) − (4)(0)) − j((3)(0) − (2)(0))
+ k((3)(4) − (2)(2)) = 8k
V × U =
∣∣∣∣∣i j k2 4 03 2 0
∣∣∣∣∣ = i((4)(0) − (2)(0)) − j((2)(0) − (3)(0))
+ k((2)(2) − (3)(4)) = −8k
Problem 2.130 What is the cross product r × F ofthe position vector r = 2i + 2j + 2k (m) and the forceF = 20i − 40k (N)?
Solution: Use Eq. (2.34) and expand into 2 by 2 determinants.
r × F =
∣∣∣∣∣i j k2 2 220 0 −40
∣∣∣∣∣ = i((2)(−40) − (0)(2)) − j((2)(−40)
− (20)(2)) + k((2)(0) − (2)(20))
r × F = −80i + 120j − 40k (N-m)
Problem 2.131 Determine the cross product r×F of
the position vector r = 4i − 12j + 3k (m) and the force
F = 16i − 22j − 10k (N).
Solution:
r × F =
∣∣∣∣∣i j k4 −12 3
16 −22 −10
∣∣∣∣∣r × F = (120 − (−66))i + (48 − (−40))j
+ (−88 − (−192))k (N-m)
r × F = 186i + 88j + 104k (N-m)
Problem 2.132 Consider the vectors U = 6i−2j−3kand V = −12i + 4j + 6k.(a) Determine the cross product U × V.(b) What can you conclude about U and V from the
result of (a)?
Solution: For (a) Use Eq. (2.34) and expand into 2 by 2 determi-nants.
U × V =
∣∣∣∣∣i j k6 −2 −3
−12 4 6
∣∣∣∣∣= i((−2)(6) − (4)(−3)) + j((6)(6)
− (−12)(−2)) + k((6)(4) − (−12)(−2))
U × V = 0i + 0j + 0k
(b) From the definition of the cross product (see Eq. (2.28)) U × V =|U||V| sin θe, where θ is the angle between the two vectors, and e is a unitvector perpendicular to both U and V. If U × V = 0 and if |U| �= 0 and|V| �= 0 then since by definition e �= 0, sin θ must be zero: sin θ = 0, andθ = 0◦ or θ = 180◦, and the two vectors are said to be parallel. (A graphicalconstruction confirms this interpretation.)
Full file at https://fratstock.euProblem 2.133 The cross product of two vectors Uand V is U × V = −30i + 40k. The vector V =4i − 2j + 3k. Determine the components of U.
Solution: We know
U × V =
∣∣∣∣∣i j k
Ux Uy Uz
4 −2 3
∣∣∣∣∣U × V = (3Uy + 2Uz)i + (4Uz − 3Ux)j + (−2Ux − 4Uy)k (1)
We also know
U × V = −30i + 0j + 40k (2)
Equating components of (1) and (2), we get
3Uy + 2Uz = −30
4Uz − 3Ux = 0
−2Ux − 4Uy = 40
Setting Ux = 4 and solving, we get
U = 4i − 12j + 3k
Problem 2.134 The magnitudes |U| = 10 and |V| =20.(a) Use the definition of the cross product to determine
U × V.(b) Use the definition of the cross product to determine
V × U.(c) Use Eq. (2.34) to determine U × V.(d) Use Eq. (2.34) to determine V × U.
U
V
x
y
45°30°
Solution: From Eq. (228) U × V = |U||V| sin θe. From thesketch, the positive z-axis is out of the paper. For U × V, e = −1k(points into the paper); for V×U, e = +1k (points out of the paper).The angle θ = 15◦, hence (a) U × V = (10)(20)(0.2588)(e) =51.8e = −51.8k. Similarly, (b) V × U = 51.8e = 51.8k (c) Thetwo vectors are:
U = 10(i cos 45◦ + j sin 45) = 7.07i + 0.707j,
V = 20(i cos 30◦ + j sin 30◦) = 17.32i + 10j
U × V =
∣∣∣∣∣i j k
7.07 7.07 017.32 10 0
∣∣∣∣∣ = i(0) − j(0) + k(70.7 − 122.45)
= −k51.8
(d) V × U =
∣∣∣∣∣i j k
17.32 10 07.07 7.07 0
∣∣∣∣∣ = i(0) − j(0) + k(122.45 − 70.7)
= 51.8k
y
x
VU
45°
30°
Full file at https://fratstock.euProblem 2.135 The force F = 10i − 4j (N). Deter-mine the cross product rAB × F. y
x
B
A
rAB
(6, 3, 0) m
(6, 0, 4) m
F
z
Solution: The position vector is
rAB = (6 − 6)i + (0 − 3)j + (4 − 0)k = 0i − 3j + 4k
The cross product:
rAB × F =
∣∣∣∣∣i j k0 −3 410 −4 0
∣∣∣∣∣ = i(16) − j(−40) + k(30)
= 16i + 40j + 30k (N-m)
y
x
B
A
rAB
(6, 3, 0)
(6, 0, 4) Fz
Problem 2.136 By evaluating the cross product U ×V, prove the identity sin(θ1 − θ2) = sin θ1 cos θ2 −cos θ1 sin θ2.
x
y
V
θ1
θ2
U
Solution: Assume that both U and V lie in the x-y plane. Thestrategy is to use the definition of the cross product (Eq. 2.28) andthe Eq. (2.34), and equate the two. From Eq. (2.28) U × V =|U||V| sin(θ1 − θ2)e. Since the positive z-axis is out of the pa-per, and e points into the paper, then e = −k. Take the dot productof both sides with e, and note that k · k = 1. Thus
sin(θ1 − θ2) = −(
(U × V) · k|U||V|
)
The vectors are:
U = |U|(i cos θ1 + j sin θ2), and V = |V|(i cos θ2 + j sin θ2).
The cross product is
U × V =
∣∣∣∣∣i j k
|U| cos θ1 |U| sin θ1 0|V| cos θ2 |V| sin θ2 0
∣∣∣∣∣= i(0) − j(0) + k(|U||V|)(cos θ1 sin θ2 − cos θ2 sin θ1)
Substitute into the definition to obtain: sin(θ1−θ2) = sin θ1 cos θ2−cos θ1 sin θ2. Q.E.D.
y
x
U
V
θ
θ1
2
Full file at https://fratstock.euProblem 2.137 Use the cross product to determine thecomponents of a unit vector e that is normal to both ofthe vectors U = 8i − 6j + 4k and V = 3i + 7j + 9k.
Solution: First, find U × V = R
R = U × V =
∣∣∣∣∣i j k8 −6 43 7 9
∣∣∣∣∣R = (−54 − 28)i + (12 − 72)j + (56 − (−18)) k
R = −82i − 60j + 74k
eR = ± R|R| = ±
(−82i − 60j + 74k125.7
)
er = ±(−0.652i − 0.477j + 0.589k)
Problem 2.138 (a) What is the cross product rOA ×rOB? (b) Determine a unit vector e that is perpendicularto rOA and rOB .
A (6, –2, 3) m
(4, 4, –4) mB
x
y
z
O
rOB
rOA
Solution: The two radius vectors are
rOB = 4i + 4j − 4k, rOA = 6i − 2j + 3k
(a) The cross product is
rOA × rOB =
∣∣∣∣∣i j k6 −2 34 4 −4
∣∣∣∣∣ = i(8 − 12) − j(−24 − 12)
+ k(24 + 8)
= −4i + 36j + 32k (m2)
The magnitude is
|rOA × rOB | =√
42 + 362 + 322 = 48.33 m2
(b) The unit vector is
e = ±(
rOA × rOB
|rOA × rOB |)
= ±(−0.0828i + 0.7448j + 0.6621k)
(Two vectors.)
A(6, –2, 3)
(4, 4, –4)B
x
y
z
O
rOB
rOA
Full file at https://fratstock.euProblem 2.139 For the points O, A, and B in Pro-blem 2.138, use the cross product to determine the lengthof the shortest straight line from point B to the straightline that passes through points O and A.
Solution:
rOA = 6i − 2j + 3k (m)
rOB = 4i + 4j − 4k (m)
rOA × rOB = C
(C is ⊥ to both rOA and rOB)
C =
∣∣∣∣∣i j k6 −2 34 4 −4
∣∣∣∣∣ =(+8 − 12)i+(12 + 24)j+(24 + 8)k
C = −4i + 36j + 32k
C is ⊥ to both rOA and rOB . Any line ⊥ to the plane formed by Cand rOA will be parallel to the line BP on the diagram. C × rOA
is such a line. We then need to find the component of rOB in thisdirection and compute its magnitude.
C × rOA =
∣∣∣∣∣i j k
−4 +36 326 − 2 3
∣∣∣∣∣C = 172i + 204j − 208k
The unit vector in the direction of C is
eC =C|C| = 0.508i + 0.603j − 0.614k
(The magnitude of C is 338.3)We now want to find the length of the projection, P , of line OB in direction ec.
P = rOB · eC
= (4i + 4j − 4k) · eC
P = 6.90 m
A(6, –2, 3) m
(4, 4, –4) mB
x
y
z
O
rOB
rOA
P
Problem 2.140 The cable BC exerts a 1000-lb forceF on the hook at B. Determine rAB × F.
rAB
A
B
C
z
x
y
rAC
F6 ft
8 ft
4 ft
4 ft 12 ft
Solution: The coordinates of points A, B, and C are A (16, 0,12), B (4, 6, 0), C (4, 0, 8). The position vectors are
rOA = 16i + 0j + 12k, rOB = 4i + 6j + 0k, rOC = 4i + 0j + 8k.
The force F acts along the unit vector
eBC =rBC
|rBC | =rOC − rOB
|rOC − rOB | =rAB
|rAB |Noting rOC −rOB = (4−4)i+(0−6)j+(8−0)k = 0i−6j+8k|rOC − rOB | =
√62 + 82 = 10. Thus
eBC = 0i − 0.6j + 0.8k, and F = |F|eBC = 0i − 600j + 800k (lb).
The vector
rAB = (4 − 16)i + (6 − 0)j + (0 − 12)k = −12i + 6j − 12k
Thus the cross product is
rAB × F =
∣∣∣∣∣i j k
−12 6 −120 −600 800
∣∣∣∣∣ = −2400i + 9600j + 7200k (ft-lb)
r
A
B
C
x
y
6 ft
8 ft
4 ft
4 ft 12 ft
Full file at https://fratstock.euProblem 2.141 The cable BC shown in Prob-lem 2.140 exerts a 300-lb force F on the hook at B.(a) Determine rAB × F and rAC × F.(b) Use the definition of the cross product to explain
why the result of (a) are equal.
Solution: (a) From Problem 2.140, the unit vector
eBC = 0i − 0.6j + 0.8k, and rAB = −12i + 6j − 12k
Thus F = |F|eBC = 0i − 180j + 240k, and the cross product is
rAB × F =
∣∣∣∣∣i j k
−12 6 −120 −180 240
∣∣∣∣∣ = −720i + 2880j + 2160k (ft-lb)
The vector rAC = (4 − 16)i+ 0j+ (8 − 12)k = −12i+ 0j− 4k.Thus the cross product is
rAC × F =
∣∣∣∣∣i j k
−12 0 −40 −180 240
∣∣∣∣∣ = −720i + 2880j + 2160k (ft-lb)
(b) The definition of the cross product is r × F = |r||F| sin θe.Since the two cross products above are equal, |rAB ||F| sin θ1e =|rAC ||F| sin θ2e. Note that rAC = rAB + rBC from Prob-lem 2.116, hence rAC × F = rAB × F + rBC × F =|rAB ||F| sin θ1e + |rBC ||F| sin 0e = |rAB ||F| sin θ1e, sincerBC and F are parallel. Thus the two results are equal.
Problem 2.142 The rope AB exerts a 50-N force T onthe collar at A. Let rCA be the position vector from pointC to point A. Determine the cross product rCA × T.
0.4 m
0.5 m
0.15 m
0.3 m0.2 m
0.25 m
0.2 mz
x
y
A
B C
D
O
T
Solution: The vector from C to D is rCD = (xD − xC)i +(yD − yC)j + (zD − zC)k. The magnitude of the vector
|rCD| =√
(xD − xC)2 + (yD − yC)2 + (zD − zC)2.
The components of the unit vector along CD are given by uCDx =(xD − xC)/|rCD|, uCDy = (yD − yC)/|rCD|, etc. Numericalvalues are |rCD| = 0.439 m, uCDx = −0.456, uCDy = −0.684,and uCDz = 0.570. The coordinates of point A are given by xA =xC + |rCA|uCDx, yA = yC + |rCA|uCDy , etc. The coordinatesof point A are (0.309, 0.162, 0.114) m. The vector rCA is given byrCA = (xA−xC)i+(yA−yC)j+(zA−zC)k. The vector rCA isrCA = (−0.091)i + (−0.137)j + (0.114)k m. The vector from Ato B and the corresponding unit vector are found in the same manneras from C to D above. The results are |rAB | = 0.458 m, uABx =−0.674, uABy = 0.735, and uABz = 0.079. The force T is givenby T = |T|uAB . The result is T = −33.7i + 36.7j + 3.93k N.The cross product rCA × T can now be calculated.
rCA × T =
∣∣∣∣∣i j k
−0.091 −0.138 0.114−33.7 36.7 3.93
∣∣∣∣∣= (−4.65)i + (−3.53)j + (−7.98)k N-m
0.4 m
0.5 m
0.15 m
0.3 m0.2 m
0.25 m
0.2 mz
x
y
BC
A
D
O
T
Full file at https://fratstock.euProblem 2.143 In Problem 2.142, let rCB be the po-sition vector from point C to point B. Determine thecross product rCB ×T and compare your answer to theanswer to Problem 2.142.
Solution: We need rCB and T in component form.
rCB = (xB − xC)i + (yB − yC)j + (zB − zC)k
where B is at (0, 0.5, 0.15) (m) and C is at (0.4, 0.3, 0) (m)
rCB = −0.4i + 0.2j + 0.15k (m)
We now need to find T . From Problem 2.142, its magnitude is 50 N.We need a unit vector eAB to be able to write T as T = 50 eAB
and then perform the required cross product. We need the coordinatesof point A. Let us find eCA = eCD and use this plus the knownlocation of C to get the location of A. Point D is located at (0.2, 0,0.25)
eCD = eCA =rCD
|rCD| = 0
eCD = −0.456i − 0.684j + 0.570k
From the diagram, dAC = 0.2 m
xA = xC + dAC(eCDx)
yA = yC + dAC(eCDy)
zA = zC + dAC(eCDz)
Recall C is at (0.4, 0.3, 0)Substituting, we find A is at (0.309, 0.163, 0.114).We now need the unit vector from A to B.
eAB =rAB
|rAB | =(xB − xA)i + (yB − yA)j + (zB − zA)k
|rAB |or
eAB = −0.674i + 0.735j + 0.078k
We now want T = |T|eAB = 50 eAB we get
T = 33.69i + 36.74j + 3.93k (N)
we can now form rCB × T
rCB × T =
∣∣∣∣∣i j k
−0.4 +0.2 +0.1533.69 36.74 3.93
∣∣∣∣∣rCB × T = −4.72i + 6.626j − 21.434k (N-m)
0.4 m
0.5 m
0.15 m
0.3 m0.2 m
0.25 m
0.2 mz
x
y
B C
A
D
O
T
Full file at https://fratstock.euProblem 2.144 The bar AB is 6 m long and is perpen-dicular to the bars AC and AD. Use the cross productto determine the coordinates xB , yB , zB of point B.
C
A
B
(0, 0, 3) m (4, 0, 0) m x
y
(0, 3, 0) m
(xB, yB, zB)
D
z
Solution: The strategy is to determine the unit vector perpendic-ular to both AC and AD, and then determine the coordinates that willagree with the magnitude of AB. The position vectors are:
rOA = 0i + 3j + 0k, rOD = 0i + 0j + 3k, and
rOC = 4i + 0j + 0k. The vectors collinear with the bars are:
rAD = (0 − 0)i + (0 − 3)j + (3 − 0)k = 0i − 3j + 3k, rAC
= (4 − 0)i + (0 − 3)j + (0 − 0)k = 4i − 3j + 0k.
The vector collinear with rAB is
R = rAD × rAC =
∣∣∣∣∣i j k0 −3 34 −3 0
∣∣∣∣∣ = 9i + 12j + 12k
The magnitude |R| = 19.21 (m). The unit vector is
eAB =R|R| = 0.4685i + 0.6247j + 0.6247k.
Thus the vector collinear with AB is
rAB = 6eAB = +2.811i + 3.75j + 3.75k.
Using the coordinates of point A:
xB = 2.81 + 0 = 2.81 (m)
yB = 3.75 + 3 = 6.75 (m)
zB = 3.75 + 0 = 3.75 (m)
x
y B
Cz
D
A[0,3,0]
[4,0,0]
[0,0,3]
Full file at https://fratstock.euProblem 2.145 Determine the minimum distancefrom point P to the plane defined by the three pointsA, B, and C.
A
(3, 0, 0) m
(0, 5, 0) mB
x
y
z
C(0, 0, 4) m
P(9, 6, 5) m
Solution: The strategy is to find the unit vector perpendicular tothe plane. The projection of this unit vector on the vector OP : rOP ·eis the distance from the origin to P along the perpendicular to the plane.The projection on e of any vector into the plane (rOA · e, rOB · e, orrOC · e) is the distance from the origin to the plane along this sameperpendicular. Thus the distance of P from the plane is
d = rOP · e − rOA · e.
The position vectors are: rOA = 3i, rOB = 5j, rOC = 4k andrOP = 9i + 6j + 5k. The unit vector perpendicular to the plane isfound from the cross product of any two vectors lying in the plane.Noting: rBC = rOC − rOB = −5j + 4k, and rBA = rOA −rOB = 3i − 5j. The cross product:
rBC × rBA =
∣∣∣∣∣i j k0 −5 43 −5 0
∣∣∣∣∣ = 20i + 12j + 15k.
The magnitude is |rBC × rBA| = 27.73, thus the unit vector ise = 0.7212i + 0.4327j + 0.5409k. The distance of point P fromthe plane is d = rOP ·e−rOA ·e = 11.792−2.164 = 9.63 m. Thesecond term is the distance of the plane from the origin; the vectorsrOB , or rOC could have been used instead of rOA.
y
z
x
B[0,5,0]
P[9,6,5]
A[3,0,0]
C[0,0,4]
O
Problem 2.146 Consider vectors U = 3i− 10j, V =−6j + 2k, and W = 2i + 6j − 4k.(a) Determine the value of the mixed triple product U ·
(V ×W) by first evaluating the cross product V ×W and then taking the dot product of the result withthe vector U.
(b) Determine the value of the mixed triple product U ·(V × W) by using Eq. (2.36).
Solution: (a) The cross product
V × W =
∣∣∣∣∣i j k0 −6 22 6 −4
∣∣∣∣∣ = (+24 − 12)i − (0 − 4)j + (0 + 12)k
= 12i + 4j + 12k
Take the dot product: U·(V×W) = (3)(12)+(4)(−10)+0 = −4(b) Eq. (2.36) expresses the mixed triple product as a 3X3 determinant.
U · (V × W)=
∣∣∣∣∣3 −10 00 −6 22 6 −4
∣∣∣∣∣ =(3)(24 − 12) − (−10)(−4) + (0)
= 36 − 40 = −4
Full file at https://fratstock.euProblem 2.147 For the vectors U = 6i + 2j − 4k,V = 2i + 7j, and W = 3i + 2k, evaluate the followingmixed triple products: (a) U·(V×W); (b) W·(V×U);(c) V · (W × U).
Solution: Use Eq. (2.36).
(a) U · (V × W) =
∣∣∣∣∣6 2 −42 7 03 0 2
∣∣∣∣∣= 6(14) − 2(4) + (−4)(−21) = 160
(b) W · (V × U) =
∣∣∣∣∣3 0 22 7 06 2 −4
∣∣∣∣∣= 3(−28) − (0) + 2(4 − 42) = −160
(c) V · (W × U) =
∣∣∣∣∣2 7 03 0 26 2 −4
∣∣∣∣∣= 2(−4) − 7(−12 − 12) + (0) = 160
Problem 2.148 Use the mixed triple product to calcu-late the volume of the parallelepiped.
x
y
z
(140, 90, 30) mm
(200, 0, 0) mm
(160, 0, 100) mm
Solution: We are given the coordinates of point D. From thegeometry, we need to locate points A and C. The key to doing thisis to note that the length of side OD is 200 mm and that side ODis the x axis. Sides OD, AE, and CG are parallel to the x axisand the coordinates of the point pairs (O and D), (A and E), and(C and D) differ only by 200 mm in the x coordinate. Thus, thecoordinates of point A are (−60, 90, 30) mm and the coordinates ofpoint C are (−40, 0, 100) mm. Thus, the vectors rOA, rOD , andrOC are rOD = 200i mm, rOA = −60i + 90j + 30k mm, andrOC = −40i+0j+100k mm. The mixed triple product of the threevectors is the volume of the parallelepiped.The volume is
rOA · (rOC × rOD) =
∣∣∣∣∣−60 90 30−40 0 100200 0 0
∣∣∣∣∣= −60(0) + 90(200)(100) + (30)(0) mm3
= 1,800,000 mm3
y
z
A
B
C
O
F
G
D
E
x
(140, 90, 30)mm
(200, 0, 0)mm
(160, 0, 100)mm
Full file at https://fratstock.euProblem 2.149 By using Eqs. (2.23) and (2.34), showthat
U · (V × W) =
∣∣∣∣∣∣
Ux Uy Uz
Vx Vy Vz
Wx Wy Wz
∣∣∣∣∣∣
.
Solution: One strategy is to expand the determinant in terms ofits components, take the dot product, and then collapse the expan-sion. Eq. (2.23) is an expansion of the dot product: Eq. (2.23):U · V = UXVX + UY VY + UZVZ . Eq. (2.34) is the determi-nant representation of the cross product:
Eq. (2.34) U × V =
∣∣∣∣∣i j k
UX UY UZ
VX VY VZ
∣∣∣∣∣For notational convenience, write P = (U × V). Expand the deter-minant about its first row:
P = i
∣∣∣∣ UY UZ
VY VZ
∣∣∣∣ − j
∣∣∣∣ UX UZ
VX VZ
∣∣∣∣ + k
∣∣∣∣ UX UZ
VX VZ
∣∣∣∣
Since the two-by-two determinants are scalars, this can be written in the form:P = iPX + jPY + kPZ where the scalars PX , PY , and PZ are the two-by-two determinants. Apply Eq. (2.23) to the dot product of a vector Q with P.Thus Q · P = QXPX + QY PY + QZPZ . Substitute PX , PY , and PZ intothis dot product
Q · P = QX
∣∣∣∣ UY UZ
VY VZ
∣∣∣∣ − QY
∣∣∣∣ UX UZ
VX VZ
∣∣∣∣ + Qz
∣∣∣∣ UX UZ
VX VZ
∣∣∣∣But this expression can be collapsed into a three-by-three determinant di-rectly, thus:
Q · (U × V) =
∣∣∣∣∣QX QY QZ
UX UY UZ
VX VY VZ
∣∣∣∣∣. This completes the demonstration.
Problem 2.150 The vectors U = i+UY j+4k, V =2i + j − 2k, and W = −3i + j − 2k are coplanar (theylie in the same plane). What is the component Uy?
Solution: Since the non-zero vectors are coplanar, the cross pro-duct of any two will produce a vector perpendicular to the plane, andthe dot product with the third will vanish, by definition of the dotproduct. Thus U · (V × W) = 0, for example.
U · (V × W) =
∣∣∣∣∣1 UY 42 1 −2
−3 1 −2
∣∣∣∣∣= 1(−2 + 2) − (UY )(−4 − 6) + (4)(2 + 3)
= +10UY + 20 = 0
Thus UY = −2
Problem 2.151 The magnitude of F is 8 kN. ExpressF in terms of scalar components.
F
x
y
(7, 2) m
(3, 7) m
Solution: The unit vector collinear with the forceF is developed as follows:The collinear vector is r = (7 − 3)i + (2 − 7)j = 4i − 5jThe magnitude: |r| =
√42 + 52 = 6.403 m. The unit vector is
e = r|r| = 0.6247i − 0.7809j. The force vector is
F = |F|e = 4.997i − 6.247j = 5i − 6.25j (kN)
x
y
F
(3,7)m
(7,2)m
Full file at https://fratstock.euProblem 2.152 The magnitude of the vertical forceWis 600 lb, and the magnitude of the force B is 1500 lb.Given that A + B + W = 0, determine the magnitudeof the force A and the angle α.
50°
B W
Aα
Solution: The strategy is to use the condition of force balance todetermine the unknowns. The weight vector is W = −600j. Thevector B is
B = 1500(i cos 50◦ + j sin 50◦) = 964.2i + 1149.1j
The vector A is A = |A|(i cos(180 + α) + j sin(180 + α))A = |A|(−i cos α − j sin α). The forces balance, hence A +B + W = 0, or (964.2 − |A| cos α)i = 0, and (1149.1 − 600 −|A| sin α)j = 0. Thus |A| cos α = 964.2, and |A| sin α = 549.1.Take the ratio of the two equations to obtain tan α = 0.5695, or
α = 29.7◦. Substitute this angle to solve: |A| = 1110 lb
50°B W
Aα
Problem 2.153 What are the direction cosines of F?
x
y
z
F = 20i + 10j – 10k (lb)A
(8, 1, –2) ft
(4, 4, 2) ft
B
θ
Solution: Use the definition of the direction cosines and the en-suing discussion.The magnitude of F: |F| =
√202 + 102 + 102 = 24.5.
The direction cosines are cos θx = Fx|F| = 20
24.5 = 0.8165,
cos θy =Fy
|F| =10
24.5= 0.4082
cos θz =Fz
|F| =−1024.5
= −0.4082
x
y
z
F = 20i + 10j – 10k (lb)A
(8, 1, –2) ft
(4, 4, 2) ft
B
θ
Problem 2.154 Determine the scalar components ofa unit vector parallel to line AB that points from A to-ward B.
Solution: Use the definition of the unit vector, we getThe position vectors are: rA = 4i+4j+2k, rB = 8i+1j− 2k. The vectorfrom A to B is rAB = (8 − 4)i + (1 − 4)j + (−2 − 2)k = 4i − 3j − 4k.The magnitude: |rAB | =
√42 + 32 + 42 = 6.4. The unit vector is
eAB =rAB
|rAB | =4
6.4i − 3
6.4j − 4
6.4k = 0.6247i − 0.4688j − 0.6247k
Full file at https://fratstock.euProblem 2.155 What is the angle θ between the lineAB and the force F?
Solution: Use the definition of the dot product Eq. (2.18), andEq. (2.24):
cos θ =rAB · F|rAB ||F| .
From the solution to Problem 2.130, the vector parallel to AB isrAB = 4i − 3j − 4k, with a magnitude |rAB | = 6.4. From Prob-lem 2.129, the force is F = 20i + 10j − 10k, with a magnitude of|F| = 24.5. The dot product is rAB · F = (4)(20) + (−3)(10) +(−4)(−10) = 90. Substituting, cos θ = 90
(6.4)(24.5) = 0.574,
θ = 55◦
Problem 2.156 Determine the vector component of Fthat is parallel to the line AB.
Solution: Use the definition in Eq. (2.26): UP = (e · U)e, where e isparallel to a line L. From Problem 2.130 the unit vector parallel to line AB iseAB = 0.6247i − 0.4688j − 0.6247k. The dot product is
e · F = (0.6247)(20) + (−0.4688)(10) + (−0.6247)(−10) = 14.053.
The parallel vector is
(e · F)e = (14.053)e = 8.78i − 6.59j − 8.78k (lb)
Problem 2.157 The magnitude of FB is 400 N and|FA +FB | = 900 N. Determine the components of FA.
x
y
z
60°
30° 50°
40°
FA
FB
Solution:
|FB | = 400 N
We need to write each vector in terms of its known or unknown com-ponents. From the diagram
FAx = (|FA| cos 40◦) cos 40◦ = 0.587
FAz = (|FA| cos 40◦) cos 50◦ = 0.492
FAy = |FA| sin 40◦ = 0.642
FBx = −(400 cos 60◦) cos 60◦
FBz = (400 cos 60◦) cos 30◦
FBy = 400 sin 60◦
Let FA = |FA| and FB = |FB | = 400 N.The components of the vectors are
FA = FA cos 40◦ sin 50◦i + FA sin 40◦j + FA cos 40◦ cos 50◦k
= FA(0.587i + 0.643j + 0.492k), (1)
FB = −FB cos 60◦ sin 30◦i + FB sin 60◦j + FB cos 60◦ cos 30◦k
= −100i + 346j + 173k (N).
Setting
900 N = |FA + FB |= [(0.587FA − 100)2 + (0.643FA + 346)2
+(0.492FA + 173)2]1/2
and solving, we obtain FA = 595 N. Substituting this result into Eq. (1),
FA = 349i + 382j + 293k (N).
x
y
z
60°
30°20°
50°
40°
40°
FA
FB
Full file at https://fratstock.euProblem 2.158 Suppose that the forces FA and FB
shown in Problem 2.163 have the same magnitude andFA · FB = 600 N2. What are FA and FB?
Solution: From Problem 2.163, the forces are:
FA = |FA|(i cos 40◦ sin 50◦ + j sin 40◦ + k cos 40◦ cos 50◦)
= |FA|(0.5868i + 0.6428j + 0.4924k)
FB = |FB |(−i cos 60◦ sin 30◦ + j sin 60◦ + k cos 60◦ cos 30◦)
= |FB |(−0.25i + 0.866j + 0.433k)
The dot product: FA · FB = |FA||FB |(0.6233) = 600 N2, from
|FA| = |FB | =
√600
0.6233= 31.03 N,
and
FA = 18.21i + 19.95j + 15.28k (N)
FB = −7.76i + 26.87j + 13.44k (N)
Problem 2.159 The rope CE exerts a 500-N force Ton the door ABCD. Determine the vector componentof T in the direction parallel to the line from point A topoint B.
x
y
z
A (0.5,0,0) m
E
B
C
D
T(0,0.2,0) m
(0.35,0,0.2) m
(0.4,0.25,–0.1) m
Solution: Two vectors are needed, rCE and rAB . The endpoints of these vectors are given in the figure. Thus, rCE =(xE − xC)i + (yE − yC)j + (zE − zC)k and a similar form holdsfor rAB . Calculating these vectors, we get
rCE = 0.4i + 0.05j − 0.1k m and rAB = −0.15i + 0j + 0.2k m.
The unit vector along CE is eCE = 0.963i + 0.120j − 0.241kand the force T, is T = |T|eCE . Hence, T = 500(0.963i +0.120j − 0.241k) = 482i + 60.2j − 120k N. The unit vector alongAB is given by eAB = −0.6i + 0j + 0.8k and the component ofT parallel to AB is given by TAB = T • eAB . Thus, TAB =(482)(−0.6) + (60.2)(0) + (−120)(0.8) = −385.2 N
x
y
z
A (0.5,0,0) m
E
B
C
D
T(0,0.2,0) m
(0.35,0,0.2) m
(0.4,0.25,–0.1) m
Problem 2.160 In Problem 2.169, let rBC be the po-sition vector from point B to point C. Determine thecross product rBC × T.
Solution: The vector from B to C is
rBC = (xC − xB)i + (yC − yB)j + (zC − zB)k
= −0.35i + 0.2j − 0.2k m.
The vector T is T = 482i + 60.2j − 120k N. The cross product ofthese vectors is given by
rBC × T =
∣∣∣∣∣i j k
−0.35 0.2 −0.2482 60.2 −120
∣∣∣∣∣ = −12.0i − 138j − 117k N m